This deck builds exponential functions from the ground up. It contrasts a constant rate with a constant ratio in a table, explains the restrictions on the base and why they exist, and covers growth and decay graphs with their horizontal asymptote and the transformations that move it. It then introduces the natural base e as the limit of compounding more and more often, computes compound and continuous interest with real numbers, and works through growth, decay, half-life, and cooling models, fitting a model to two data points, and solving exponential equations with a common base. It targets four killer errors: confusing a variable base with a variable exponent, multiplying the base by the exponent, forgetting that a vertical shift moves the horizontal asymptote, and using the annual interest formula when the problem says monthly.
Subject: College Algebra · 139 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
College Algebra - Deck 16
Linear growth adds the same amount every step. Exponential growth multiplies by the same amount every step. That one word - multiplies - is the whole deck.
Objectives
Money in a bank, a population of bacteria, a drug leaving your bloodstream, a hot cup of coffee - all of these change by a constant ratio, not a constant amount. That is what an exponential function describes.
Along the way we will meet four mistakes that look completely reasonable and are completely wrong. Naming them now is the cheapest insurance you will ever buy.
Warm-up
Discussion prompt
Before we open Exponential Functions and Growth: without looking back, what was the main idea of Rational Functions, Asymptotes, and Inequalities, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck covers domains, holes, and vertical, horizontal, and slant asymptotes, then sign charts and full sketches of rational functions, and finishes with polynomial and rational inequalities. It targets the classic errors: calling a cancelled factor a vertical asymptote, reading a horizontal asymptote off the constant terms, multiplying a rational inequality by a denominator whose sign is unknown, and including a zero of the denominator in a non-strict solution set.
Section
Part 1
Concept
Suppose something grows every year. There are two completely different ways it could do that.
Way one: add the same amount each year. Save 50 dollars every month. That is linear.
Way two: multiply by the same number each year. Your balance grows by 6 percent, so it gets multiplied by 1.06 each year. That is exponential.
Both feel like growth. Over a few steps they look similar. Over many steps they are not remotely the same.
Counterexample
Discussion prompt
Suppose something grows every year. There are two completely different ways it could do that.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
Both feel like growth. Over a few steps they look similar. Over many steps they are not remotely the same.
Intuition
Picture stacking bricks: one brick per step, forever. The pile grows steadily and predictably. Step one hundred is one hundred bricks tall. Nothing surprising ever happens.
Now picture a snowball rolling downhill. Each turn it picks up snow in proportion to how big it already is. The bigger it gets, the faster it gets bigger.
That feedback - the growth depending on the current size - is exactly what multiplying by a fixed number does. It is why exponential growth starts out looking boring and ends up looking impossible.
exponential growth — Growth in which the quantity is multiplied by the same fixed number over each equal step of the input. The step size of the growth is proportional to the current amount.
Analogy
Discussion prompt
Explain Adding stacks. Multiplying snowballs. by analogy to something with no College Algebra in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Picture stacking bricks: one brick per step, forever. The pile grows steadily and predictably. Step one hundred is one hundred bricks tall. Nothing surprising ever happens.
Concept
This gives you a test you can run on any table, in about fifteen seconds.
Subtract consecutive outputs. If the difference is always the same number, the table is linear.
Divide consecutive outputs. If the ratio is always the same number, the table is exponential.
| x | linear f(x) | difference | exponential g(x) | ratio |
|---|---|---|---|---|
| 0 | 3 | - | 3 | - |
| 1 | 8 | 5 | 6 | 2 |
| 2 | 13 | 5 | 12 | 2 |
| 3 | 18 | 5 | 24 | 2 |
| 4 | 23 | 5 | 48 | 2 |
The left column adds 5 each step. The right column doubles each step. By the time we reach a modest input the two are already far apart - and the gap only widens.
One warning before you use the test: the inputs must step by equal amounts. If the x column jumps 0, 1, 2, then 5, the differences and ratios are not comparable.
Pattern
Step through it
Step through Constant difference versus constant ratio one row at a time. What is driving the change, and what would the row after the last one be?
Ranking
Put in order
Put the moves of Worked example: classify a table and write the model into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Subtracting consecutive values gives negative 200, then negative 150, then negative 112.5, then negative 84.375.
Worked example
A laptop is bought for 800 dollars. Its resale value is tracked once a year. Is the value linear or exponential, and what is the model?
| years owned | value in dollars |
|---|---|
| 0 | 800 |
| 1 | 600 |
| 2 | 450 |
| 3 | 337.5 |
| 4 | 253.125 |
Check the differences first
Why: Subtracting consecutive values gives negative 200, then negative 150, then negative 112.5, then negative 84.375. Those are not equal, so the table is not linear.
Now check the ratios
Why: Divide each value by the one before it. Every single quotient comes out to 0.75, so the table has a constant ratio.
\[ \frac{600}{800} = 0.75, \quad \frac{450}{600} = 0.75, \quad \frac{337.5}{450} = 0.75, \quad \frac{253.125}{337.5} = 0.75 \]
Identify the starting value and the base
Why: The value at zero years is 800, so that is the starting amount. The constant ratio 0.75 is the base.
\[ V(t) = 800(0.75)^{t} \]
Say what the base means in words
Why: Multiplying by 0.75 keeps 75 percent of the value, so the laptop loses 25 percent of its worth each year. A base below one always means decay.
Verify by testing the model at four years
Why: Raise 0.75 to the fourth power and multiply by 800. It must reproduce the table entry exactly, and it does.
\[ V(4) = 800(0.75)^{4} = 800(0.31640625) = 253.125 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "classify a table and write the model", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Subtracting consecutive values gives negative 200, then negative 150, then negative 112.5, then negative 84.375. Those are not equal, so the table is not linear.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The mistake: reading an exponential function as if it were a power function - putting the input in the wrong slot.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The 3 and the x get swapped, producing 4 raised to the third power.
The base is locked. The exponent is the slot the input goes into.
Why: The 3 and the x get swapped, producing 4 raised to the third power.
Trap
The mistake: reading an exponential function as if it were a power function - putting the input in the wrong slot.
\[ f(x) = 3^{x}, \qquad \text{find } f(4) \]
Student substitutes 4 for the base
Why: The 3 and the x get swapped, producing 4 raised to the third power.
\[ f(4) \;\overset{?}{=}\; 4^{3} = 64 \quad \text{(wrong)} \]
It looks harmless because both numbers are present. But this quietly turns an exponential function into a cubic one, and every later answer inherits the error.
The base is locked. The exponent is the slot the input goes into.
\[ f(x) = 3^{x}, \qquad \text{find } f(4) \]
The 3 stays put; the 4 becomes the exponent
Why: In an exponential function the variable lives upstairs. The base never changes.
\[ f(4) = 3^{4} = 81 \quad \checkmark \]
Say the two families out loud until they sound different: a power function has the variable on the bottom, an exponential function has the variable on top.
\[ \underbrace{g(x) = x^{3}}_{\text{power}} \qquad \underbrace{f(x) = 3^{x}}_{\text{exponential}} \]
Notation
Annotate
From Trap: a variable base is not a variable exponent — read this one piece at a time. What is each part doing?
On: \( f(x) = 3^{x}, \qquad \text{find } f(4) \)
Ranking
Put in order
These are the steps of Pattern: linear, exponential, or neither, scrambled. Put them back in order before the next slide shows you.
Why: This is the order the recipe itself gives. Recalling the sequence without the slide in front of you is the difference between recognising the method and being able to run it — most of what goes wrong in practice is a step done out of turn.
Pattern
Run this on any table. It takes fifteen seconds and it never lies.
For an exponential table whose inputs start at zero, the output at zero is the starting value - the number out front.
\[ y = a\,b^{x} \quad\text{where } a = y(0) \text{ and } b = \text{the constant ratio} \]
Picture it
Animation
Shows: Adding versus multiplying — a rendered Manim animation.
Rendered with Manim.
Takeaway: The linear one leads early and loses permanently.
Check
Run the pattern on this table before you look at the choices.
| x | y |
|---|---|
| 0 | 6 |
| 1 | 9 |
| 2 | 13.5 |
| 3 | 20.25 |
Check your understanding
Which model fits the table exactly?
Answer: A
Why: The differences are 3, 4.5, and 6.75 - not constant, so it is not linear. The ratios are 9/6 = 1.5, 13.5/9 = 1.5, and 20.25/13.5 = 1.5 - constant, so it is exponential with base 1.5 and starting value 6. Check x = 3: 6(1.5)^3 = 6(3.375) = 20.25.
Pattern
Step through it
Step through Check yourself: which model fits? one row at a time. What is driving the change, and what would the row after the last one be?
Section
Part 2
Concept
Here is the official form. The number out front is the starting value; the number being raised to a power is the base.
\[ f(x) = a\,b^{x}, \qquad a \ne 0, \quad b > 0, \quad b \ne 1 \]
base — The fixed number that gets multiplied in once per unit step of the input. It must be positive and not equal to one.
Those two restrictions on the base are not bureaucracy. Each one rules out a case that would break the function. We will look at both.
The restriction on the number out front is easier: if it were zero, the whole function would be flat zero forever.
Definition probe
Sort into buckets
Every line below is part of the definition of exponential growth or of base — one or the other, never both. Put each where it belongs.
Concept
Try it. Raise one to any power you like.
\[ 1^{0} = 1, \quad 1^{5} = 1, \quad 1^{-3} = 1, \quad 1^{1/2} = 1 \]
One times itself, any number of times, is one. The output never moves.
\[ f(x) = a\cdot 1^{x} = a \quad \text{for every } x \]
That is a horizontal line - a constant function. It is a perfectly fine function, it just is not exponential. Nothing grows, nothing decays, so we exclude it.
Explain it
Discussion prompt
Explain Why the base cannot be one to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
One times itself, any number of times, is one. The output never moves.
Concept
Now suppose someone hands you a negative base and asks for the output at one half.
\[ f(x) = (-4)^{x} \quad\Longrightarrow\quad f\!\left(\tfrac{1}{2}\right) = (-4)^{1/2} = \sqrt{-4} \]
That is an even root of a negative number. There is no real value. So the function would have holes scattered all through it - no smooth curve, no usable graph.
Even at whole-number inputs a negative base flips sign every step, so the outputs bounce above and below the axis instead of forming a curve.
\[ (-4)^{1} = -4, \quad (-4)^{2} = 16, \quad (-4)^{3} = -64 \]
A base of zero is just as bad: zero raised to zero is undefined, and zero to a negative power means dividing by zero.
\[ 0^{-2} = \frac{1}{0^{2}} = \frac{1}{0} \quad \text{undefined} \]
Requiring a positive base that is not one is exactly what it takes to get one unbroken, always-positive curve.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The mistake: exponents are not multiplication in disguise, but under time pressure they get treated that way.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The student writes 2 to the third power as 2 times 3, which is 6.
Exponent means repeated multiplication of the base by itself. Nothing else.
Why: The student writes 2 to the third power as 2 times 3, which is 6. Repeated multiplication has been replaced by a single multiplication.
Trap
The mistake: exponents are not multiplication in disguise, but under time pressure they get treated that way.
\[ f(x) = 5\cdot 2^{x}, \qquad \text{find } f(3) \]
First slip: the exponent gets multiplied into the base
Why: The student writes 2 to the third power as 2 times 3, which is 6. Repeated multiplication has been replaced by a single multiplication.
\[ 2^{3} \;\overset{?}{=}\; 2\cdot 3 = 6 \quad \text{(wrong)} \]
Second slip: the coefficient gets pulled inside
Why: The 5 out front is dragged into the base, giving 10 raised to the third power. Nothing in the expression grouped them together.
\[ 5\cdot 2^{3} \;\overset{?}{=}\; (5\cdot 2)^{3} = 10^{3} = 1000 \quad \text{(wrong)} \]
Exponent means repeated multiplication of the base by itself. Nothing else.
\[ 2^{3} = 2\cdot 2\cdot 2 = 8 \quad \checkmark \]
Order of operations: power first, then the coefficient
Why: There are no parentheses around the 5 and the 2, so the exponent applies only to the 2. Evaluate the power, and only then multiply by 5.
\[ f(3) = 5\cdot 2^{3} = 5\cdot 8 = 40 \quad \checkmark \]
Sanity check the size: 1000 would be twenty-five times too big. When an exponential answer feels enormous, ask whether something got multiplied that should have been raised.
Translation
\( 5\cdot 2^{3} \;\overset{?}{=}\; (5\cdot 2)^{3} = 10^{3} = 1000 \quad \text{(wrong)} \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Concept
Only one number in the whole function decides whether the curve rises or falls: the base.
If the base is greater than one, each step multiplies by something bigger than one, so the output climbs. That is growth.
\[ b > 1 \;\Longrightarrow\; \text{growth} \]
If the base is between zero and one, each step multiplies by a proper fraction, so the output shrinks. That is decay.
\[ 0 < b < 1 \;\Longrightarrow\; \text{decay} \]
| x | growth: 2 to the x | decay: one half to the x |
|---|---|---|
| -2 | 1/4 | 4 |
| -1 | 1/2 | 2 |
| 0 | 1 | 1 |
| 1 | 2 | 1/2 |
| 2 | 4 | 1/4 |
| 3 | 8 | 1/8 |
Read the two output columns side by side: they are mirror images. Decay with a base of one half is growth with a base of two, run backwards.
Pattern
Step through it
Step through Growth or decay is decided by the base alone one row at a time. What is driving the change, and what would the row after the last one be?
Picture it
Figure (svg): Graph of y equals 2 to the x, hugging the horizontal axis on the left, passing through the point (0, 1), and rising steeply to the right.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Graph the basic doubling function and list every feature a test would ask for.
Worked example
Graph the basic doubling function and list every feature a test would ask for.
\[ f(x) = 2^{x} \]
Build a small table using both negative and positive inputs
Why: Negative inputs are the part people skip, and they are exactly the part that shows the curve flattening toward the axis.
| x | f(x) |
|---|---|
| -3 | 1/8 |
| -2 | 1/4 |
| -1 | 1/2 |
| 0 | 1 |
| 1 | 2 |
| 2 | 4 |
| 3 | 8 |
Plot the points and connect them with a smooth curve
Why: The curve hugs the horizontal axis on the left and shoots upward on the right. It never touches the axis and never turns around.
Figure (svg): Graph of y equals 2 to the x, hugging the horizontal axis on the left, passing through the point (0, 1), and rising steeply to the right.
Read off the four standard features
Why: Every basic exponential has the same four: all real inputs are allowed, all outputs are positive, the curve crosses the vertical axis at the coefficient, and the horizontal axis is the asymptote.
\[ \text{domain } (-\infty, \infty), \quad \text{range } (0, \infty), \quad y\text{-intercept } (0,1), \quad \text{asymptote } y = 0 \]
Verify a point on the curve and the constant ratio
Why: Substituting 3 gives 8, which is where the plotted curve sits, and dividing each table output by the one before it gives 2 every time - the doubling the base promised.
\[ f(3) = 2^{3} = 8 \;\checkmark \qquad \frac{1/4}{1/8} = \frac{1/2}{1/4} = \frac{1}{1/2} = \frac{2}{1} = 2 \;\checkmark \]
Pattern
Step through it
Step through Worked example: graph the growth function one row at a time. What is driving the change, and what would the row after the last one be?
Concept
Look at the left half of that graph again. As the input marches off to the left, the outputs shrink: one half, one quarter, one eighth, one sixteenth.
They get as close to zero as you like. They never reach zero, because you cannot multiply a positive number by a positive number and land on zero.
horizontal asymptote — A horizontal line the graph approaches without ever touching as the input runs off toward infinity in one direction. For the basic exponential it is the line y = 0.
\[ \text{As } x \to -\infty, \quad 2^{x} \to 0 \quad \text{but } 2^{x} \ne 0 \]
This is also why the range is only the positive numbers. An exponential with a positive coefficient never produces zero and never produces a negative.
Picture it
Animation
Shows: Exponentials have a floor, not a wall — a rendered Manim animation.
Rendered with Manim.
Takeaway: Which is why an exponential can never equal zero.
Intuition
Stand a step away from a wall. Each second, move half of the remaining distance. Half, then a quarter, then an eighth.
You will get absurdly close - a millimeter, a hair, an atom. You will never actually arrive, because there is always a remaining half to cover.
That wall is the asymptote. The graph behaves exactly like you do: closer and closer, forever, never touching.
Practical upshot on a test: an asymptote is a line you draw dashed and never cross. It is also the boundary of the range.
Concept
Every exponential passes through one point you can find without any work at all.
\[ f(0) = a\,b^{0} = a\cdot 1 = a \]
Because any allowed base raised to the zero power is one, the output at zero is just the number out front.
So the graph of the basic function crosses the vertical axis at one, and scaling by a coefficient slides that crossing to the coefficient's value.
\[ y = b^{x} \text{ passes through } (0,1); \qquad y = a\,b^{x} \text{ passes through } (0,a) \]
That single point plus the asymptote plus the direction of the curve is usually enough to sketch a decent graph by hand.
Step zero
Discussion prompt
Worked example: features of a decay function — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Classify by the base
Answer:
Worked example
Describe the graph completely: direction, intercept, asymptote, domain, and range.
\[ g(x) = 2\left(\tfrac{1}{3}\right)^{x} \]
Classify by the base
Why: The base is one third, which sits strictly between zero and one, so each step multiplies by a proper fraction. The graph decays from left to right.
Find the y-intercept from the coefficient
Why: Any allowed base to the zero power is one, so the output at zero is the coefficient 2. No computation beyond that is needed.
\[ g(0) = 2\left(\tfrac{1}{3}\right)^{0} = 2(1) = 2 \;\Rightarrow\; (0, 2) \]
Build a short table around the intercept
Why: Negative inputs invert the fraction, which is why the left side climbs steeply. Positive inputs shrink by a factor of three each step.
| x | g(x) |
|---|---|
| -2 | 18 |
| -1 | 6 |
| 0 | 2 |
| 1 | 2/3 |
| 2 | 2/9 |
State the asymptote, domain, and range
Why: No shift has been applied, so the asymptote is still the horizontal axis. Every real input is legal, and the outputs stay strictly positive.
\[ \text{asymptote } y = 0, \quad \text{domain } (-\infty, \infty), \quad \text{range } (0, \infty) \]
Verify the table entry at negative one
Why: A negative exponent flips the fraction, turning one third into three. Two times three is six, exactly the table value, so the model and the table agree.
\[ g(-1) = 2\left(\tfrac{1}{3}\right)^{-1} = 2\cdot 3 = 6 \;\checkmark \]
Picture it
Animation
Shows: Decay is the same graph, reflected — a rendered Manim animation.
Rendered with Manim.
Takeaway: A base between zero and one falls toward, but never reaches, zero.
Elimination
Eliminate the wrong options
Which function models exponential DECAY?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Decay is decided by the base being strictly between 0 and 1. In choice A the base is 0.92, so each step keeps 92 percent of the amount and the output shrinks: 4, then 3.68, then about 3.39.
Check
Only one number decides this. Find it in each choice before you answer.
Check your understanding
Which function models exponential DECAY?
Answer: A
Why: Decay is decided by the base being strictly between 0 and 1. In choice A the base is 0.92, so each step keeps 92 percent of the amount and the output shrinks: 4, then 3.68, then about 3.39.
Section
Part 3
Picture it
Figure (svg): Graph of y equals 2 to the x plus 3, flattening toward a dashed horizontal line at y equals 3 on the left and rising steeply on the right.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Adding a number on the outside of the function lifts the entire picture straight up. Every point rises by the same amount.
Concept
Adding a number on the outside of the function lifts the entire picture straight up. Every point rises by the same amount.
\[ y = 2^{x} \quad\longrightarrow\quad y = 2^{x} + 3 \]
Here is the part people forget: the asymptote is part of the picture. If the curve rises three units, so does the line it is approaching.
Figure (svg): Graph of y equals 2 to the x plus 3, flattening toward a dashed horizontal line at y equals 3 on the left and rising steeply on the right.
So the range moves too. The outputs are no longer everything above zero - they are everything above three.
\[ \text{asymptote } y = 3, \qquad \text{range } (3, \infty) \]
Subtracting on the outside does the same thing downward. The asymptote goes with it, and it can end up below the horizontal axis.
\[ y = 2^{x} - 7 \;\Longrightarrow\; \text{asymptote } y = -7, \quad \text{range } (-7, \infty) \]
Intuition
Think of the graph and its asymptote as one object drawn on a sheet of clear plastic - the curve and the dashed line together.
A vertical shift slides the whole sheet up or down. You cannot slide the curve and leave the dashed line behind; they are printed on the same sheet.
A horizontal shift slides the sheet left or right. Now the dashed line does not appear to move at all, because a horizontal line looks identical after sliding sideways.
That is the whole rule: only up-and-down motion changes where the asymptote sits. Left-and-right motion leaves it alone.
And remember the direction quirk from the transformations deck - a change inside the function moves the graph the opposite way from the sign you see.
\[ y = 2^{\,x-1} \text{ shifts RIGHT one unit} \]
Estimation
Predict first
List the transformations, the asymptote, the intercept, the domain, and the range.
Commit before you compute: what does Worked example: describe a shifted exponential come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the asymptote claim with a far-left input
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. At an input of negative ten the power is a tiny positive fraction, so the output sits a hair above three and never at or below it - exactly what an asymptote at three means.
Worked example
List the transformations, the asymptote, the intercept, the domain, and the range.
\[ g(x) = 2^{\,x-1} + 3 \]
Name the parent function
Why: Strip away the added and subtracted numbers and you are left with the doubling function, so every feature starts from that shape.
\[ f(x) = 2^{x} \]
Read the inside change: shift right 1 unit
Why: The minus one sits with the x, inside the exponent, so it is a horizontal shift - and inside changes move opposite to their sign, so minus one means right.
Read the outside change: shift up 3 units
Why: The plus three is added after the power is computed, so it lifts every output by three. That is the shift that drags the asymptote along.
State the asymptote and the range
Why: The parent asymptote y equals 0 rose by three. The outputs of the power are always positive, so adding three keeps everything strictly above three.
\[ \text{asymptote } y = 3, \qquad \text{domain } (-\infty,\infty), \qquad \text{range } (3, \infty) \]
Find the y-intercept by substituting zero
Why: The exponent becomes negative one, which flips the base into a fraction. Half plus three is three and a half.
\[ g(0) = 2^{-1} + 3 = \tfrac{1}{2} + 3 = 3.5 \;\Rightarrow\; (0, 3.5) \]
Plot two easy anchor points
Why: Choosing inputs that make the exponent 0 and 1 keeps the arithmetic clean and pins the curve down on the right-hand side.
| x | exponent | g(x) |
|---|---|---|
| 1 | 0 | 4 |
| 2 | 1 | 5 |
| 3 | 2 | 7 |
Verify the asymptote claim with a far-left input
Why: At an input of negative ten the power is a tiny positive fraction, so the output sits a hair above three and never at or below it - exactly what an asymptote at three means.
\[ g(-10) = 2^{-11} + 3 = \tfrac{1}{2048} + 3 = 3.00048828125 \;\checkmark \]
Picture it
Animation
Shows: Shifting an exponential moves its asymptote — a rendered Manim animation.
Rendered with Manim.
Takeaway: A vertical shift drags the horizontal asymptote with it.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The mistake: memorizing that an exponential's asymptote is the horizontal axis, and then never checking whether the graph was shifted.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The rule was memorized without the condition attached to it, so the plus five is treated as decoration.
The outside constant is a vertical shift, and vertical shifts move the asymptote.
Why: The rule was memorized without the condition attached to it, so the plus five is treated as decoration.
Trap
The mistake: memorizing that an exponential's asymptote is the horizontal axis, and then never checking whether the graph was shifted.
\[ h(x) = 2^{x} + 5 \]
Student reports the parent's asymptote and range
Why: The rule was memorized without the condition attached to it, so the plus five is treated as decoration.
\[ \text{asymptote } y = 0, \quad \text{range } (0, \infty) \quad \text{(wrong)} \]
Test the claim and watch it fail
Why: If the range really started at zero there would be inputs producing outputs near zero. There are none - the outputs bottom out near five.
| x | the power | h(x) |
|---|---|---|
| -10 | 0.0009765625 | 5.0009765625 |
| -20 | about 0.00000095 | about 5.00000095 |
The outside constant is a vertical shift, and vertical shifts move the asymptote.
\[ h(x) = 2^{x} + 5 \]
Move the asymptote up by the same 5
Why: The parent line y equals 0 is lifted five units along with every point on the curve.
\[ \text{asymptote } y = 5 \]
Rebuild the range from the shifted asymptote
Why: The power is always strictly positive, so adding five gives outputs strictly greater than five, with no upper limit.
\[ 2^{x} > 0 \;\Longrightarrow\; 2^{x} + 5 > 5 \;\Longrightarrow\; \text{range } (5, \infty) \;\checkmark \]
Comparison
Comparison matrix
From Trap: leaving the asymptote at zero: refill the h(x) column from what you know. The rest of the table is as it appeared.
| x | the power | h(x) |
|---|---|---|
| -10 | 0.0009765625 | 5.0009765625 |
| -20 | about 0.00000095 | about 5.00000095 |
Concept
A minus sign in front of the whole function flips the graph over the horizontal axis. Every output changes sign.
\[ y = -\,2^{x} \;\Longrightarrow\; \text{range } (-\infty, 0), \quad \text{asymptote still } y = 0 \]
A minus sign on the input flips the graph over the vertical axis instead. Left and right trade places.
\[ y = 2^{-x} = \left(\tfrac{1}{2}\right)^{x} \]
That second one is worth staring at: reflecting a growth curve across the vertical axis literally turns it into a decay curve with the reciprocal base.
| change | what moves | asymptote after |
|---|---|---|
| add c outside | graph up c units | y = c |
| subtract c inside | graph right c units | unchanged |
| minus in front | flip over horizontal axis | unchanged |
| minus on the input | flip over vertical axis | unchanged |
Discrimination
Sort into buckets
Sort these by asymptote after, from memory, without looking back at Reflections: two different mirrors. Telling them apart on the spot is the skill; the table is only where the answer happens to be written down.
Missing information
Discussion prompt
This one has an actual x-intercept, which basic exponentials never do. Find every feature.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
The minus sign is attached to the power, so it reflects over the horizontal axis first. The plus eight is added afterward, lifting everything eight units.
Worked example
This one has an actual x-intercept, which basic exponentials never do. Find every feature.
\[ h(x) = -\,2^{x} + 8 \]
Separate the two transformations
Why: The minus sign is attached to the power, so it reflects over the horizontal axis first. The plus eight is added afterward, lifting everything eight units.
Track the asymptote through both moves
Why: The reflection leaves the line y equals 0 where it is, then the vertical shift lifts it to eight. The curve now approaches eight from below.
\[ \text{asymptote } y = 8 \]
Build the range from the inequality
Why: The power is always strictly positive, so its negative is always strictly negative, and adding eight keeps every output strictly below eight.
\[ 2^{x} > 0 \;\Rightarrow\; -2^{x} < 0 \;\Rightarrow\; -2^{x} + 8 < 8 \;\Rightarrow\; \text{range } (-\infty, 8) \]
Find the y-intercept
Why: Substituting zero makes the power equal to one, so the output is negative one plus eight.
\[ h(0) = -\,2^{0} + 8 = -1 + 8 = 7 \;\Rightarrow\; (0, 7) \]
Find the x-intercept by setting the output to zero
Why: Isolating the power gives a clean equal-bases problem. Eight is two cubed, so the exponent must be three.
\[ -\,2^{x} + 8 = 0 \;\Rightarrow\; 2^{x} = 8 = 2^{3} \;\Rightarrow\; x = 3 \]
Verify both intercepts in the original function
Why: Substituting 3 must give zero and substituting 0 must give seven. Both come out exactly, so the intercepts are correct and the curve really does cross the axis once.
\[ h(3) = -\,2^{3} + 8 = -8 + 8 = 0 \;\checkmark \qquad h(0) = -1 + 8 = 7 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "a reflected and shifted exponential", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Substituting 3 must give zero and substituting 0 must give seven. Both come out exactly, so the intercepts are correct and the curve really does cross the axis once.
Prediction
Predict first
What are the horizontal asymptote and the range of g?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: Asymptote y = -5; range (-5, infinity)
Why: The minus 5 is outside the power, so it shifts the whole graph down 5 and carries the asymptote from y = 0 to y = -5. Since 3 to the x is always greater than 0, subtracting 5 makes every output greater than -5, giving the range from -5 up to infinity.
Check
Find the asymptote first. The range follows from it automatically.
\[ g(x) = 3^{x} - 5 \]
Check your understanding
What are the horizontal asymptote and the range of g?
Answer: A
Why: The minus 5 is outside the power, so it shifts the whole graph down 5 and carries the asymptote from y = 0 to y = -5. Since 3 to the x is always greater than 0, subtracting 5 makes every output greater than -5, giving the range from -5 up to infinity.
Section
Part 4
Intuition
Here is a deliberately silly bank. It pays 100 percent interest per year, and you deposit one dollar.
If it pays once at the end of the year, you finish with two dollars. Obvious.
Now suppose it pays half the interest at six months, and the second half at twelve months - but the second half is computed on the larger balance. You should end up with a little more than two dollars.
\[ \left(1 + \tfrac{1}{2}\right)^{2} = (1.5)^{2} = 2.25 \]
Greedy question: what if the bank pays every month? Every day? Every second? Each split gives interest on interest sooner, so each split should give more.
It does give more - but the extra shrinks fast, and the total does not run away to infinity. It piles up against a ceiling. That ceiling is the number we call e.
Picture it
Animation
Shows: Where e comes from — a rendered Manim animation.
Rendered with Manim.
Takeaway: The limit of compounding ever more often. Nothing arbitrary about it.
Concept
Split the year into more and more equal pieces and compute the year-end balance on that one dollar each time.
\[ \left(1 + \frac{1}{n}\right)^{n} \quad \text{for larger and larger } n \]
| compounding | n | year-end balance |
|---|---|---|
| yearly | 1 | 2 |
| twice a year | 2 | 2.25 |
| quarterly | 4 | 2.44140625 |
| monthly | 12 | 2.613035 |
| daily | 365 | 2.714567 |
| hourly | 8760 | 2.718127 |
Look at what the extra money is doing. Going from yearly to quarterly gained about 44 cents. Going from daily to hourly gained about a thirtieth of a cent.
The values are converging. Push the splitting to its limit and you land on a specific irrational number.
\[ \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{n} = e \approx 2.718281828 \]
Pattern
Step through it
Step through Watch the ceiling appear one row at a time. What is driving the change, and what would the row after the last one be?
Concept
The most important thing to internalize: e is just a number, like the ratio of a circle's circumference to its diameter. It is not a variable and not an operation.
\[ e = 2.718281828459045\ldots \]
natural exponential function — The exponential function whose base is the number e. On a calculator it is the key marked e to the x. Its graph has exactly the same shape as any other growth exponential.
Because e sits between 2 and 3, its graph sits between the doubling and tripling curves. Same asymptote, same intercept, same domain and range.
\[ 2^{x} \;<\; e^{x} \;<\; 3^{x} \quad \text{for } x > 0 \]
Nothing new to learn about its graph. The reason it earns its own key is that it is the base that continuous growth naturally produces.
Picture it
Animation
Shows: The base controls the steepness — a rendered Manim animation.
Rendered with Manim.
Takeaway: A larger base climbs faster; a base below one decays.
Concept
When interest is paid a fixed number of times per year, this is the formula. Every symbol in it earns its place.
\[ A = P\left(1 + \frac{r}{n}\right)^{nt} \]
| symbol | meaning | example |
|---|---|---|
| P | principal: the amount deposited at the start | 2000 dollars |
| r | annual rate as a decimal | 0.06 for 6 percent |
| n | compounding periods per year | 12 for monthly |
| t | time in years | 5 |
| A | amount in the account at the end | what we solve for |
The two moving parts always travel together. Divide the rate by the number of periods, and multiply the years by that same number.
\[ \frac{r}{n} = \text{the rate for ONE period}, \qquad nt = \text{the NUMBER of periods} \]
That is the whole idea: chop the year into periods, charge a smaller rate each period, and do it more times.
Trade off
Comparison matrix
From The compound interest formula: every row here is a choice with a cost. Fill the example column, then say which row you would actually pick and what you give up for it.
| symbol | meaning | example |
|---|---|---|
| P | principal: the amount deposited at the start | 2000 dollars |
| r | annual rate as a decimal | 0.06 for 6 percent |
| n | compounding periods per year | 12 for monthly |
| t | time in years | 5 |
| A | amount in the account at the end | what we solve for |
Picture it
Animation
Shows: Compound interest — a rendered Manim animation.
Rendered with Manim.
Takeaway: Compounding more often converges — it does not run away.
Step zero
Discussion prompt
Worked example: compounded monthly — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Write down all four inputs before touching the formula
Answer:
Worked example
You deposit 2,000 dollars at an annual rate of 6 percent, compounded monthly. How much is in the account after 5 years?
Write down all four inputs before touching the formula
Why: Most errors in this topic are bookkeeping errors, not algebra errors. Naming the values first makes the substitution mechanical.
\[ P = 2000, \quad r = 0.06, \quad n = 12, \quad t = 5 \]
Compute the per-period rate and the number of periods
Why: Monthly means twelve periods a year, so each month earns one twelfth of the annual rate, and five years contains sixty months.
\[ \frac{r}{n} = \frac{0.06}{12} = 0.005, \qquad nt = 12(5) = 60 \]
Substitute and simplify the parenthesis first
Why: Order of operations: finish everything inside the parentheses before applying the exponent. One plus 0.005 is 1.005.
\[ A = 2000(1 + 0.005)^{60} = 2000(1.005)^{60} \]
Raise to the sixtieth power, then multiply
Why: The growth factor over the whole five years is about 1.3489, meaning the balance grows by nearly 35 percent. Only now do we round to cents.
\[ (1.005)^{60} = 1.348850153\ldots \;\Longrightarrow\; A = 2000(1.348850153) = 2697.700\ldots \]
State the answer in the form the question asked for
Why: Money answers are rounded to the nearest cent at the very end, never partway through.
\[ A \approx 2697.70 \text{ dollars} \]
Verify by a completely different route
Why: Compound one year at a time instead. Twelve months of 0.5 percent gives an effective annual factor of 1.0616778, and raising that to the fifth power reproduces the same sixty-month factor to eight decimal places.
\[ (1.005)^{12} = 1.0616778, \quad (1.0616778)^{5} = 1.34885015 \;\Rightarrow\; 2697.70 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "compounded monthly", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Compound one year at a time instead. Twelve months of 0.5 percent gives an effective annual factor of 1.0616778, and raising that to the fifth power reproduces the same sixty-month factor to eight decimal places.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Same account: 2,000 dollars, 6 percent, compounded monthly, 5 years. The word monthly changes two things, and it is easy to change only one.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The student correctly makes the monthly rate 0.005, then raises to the fifth power because the problem said five years.
The compounding word sets the value of one letter, and that letter appears in two places.
Why: The student correctly makes the monthly rate 0.005, then raises to the fifth power because the problem said five years. The exponent must count periods, not years.
Trap
Same account: 2,000 dollars, 6 percent, compounded monthly, 5 years. The word monthly changes two things, and it is easy to change only one.
Version one: divide the rate but leave the exponent as the years
Why: The student correctly makes the monthly rate 0.005, then raises to the fifth power because the problem said five years. The exponent must count periods, not years.
\[ A \;\overset{?}{=}\; 2000(1.005)^{5} = 2050.50 \quad \text{(wrong)} \]
Version two: ignore the word monthly entirely
Why: Falling back on the annual formula uses the full 6 percent once a year, which under-counts every month of interest-on-interest.
\[ A \;\overset{?}{=}\; 2000(1.06)^{5} = 2676.45 \quad \text{(wrong)} \]
The first answer is off by more than six hundred dollars, and the first version is not even a sensible amount for five years of growth.
The compounding word sets the value of one letter, and that letter appears in two places.
\[ A = P\left(1 + \frac{r}{\boxed{n}}\right)^{\boxed{n}t} \]
Change both places at once
Why: Monthly means n is twelve. Divide the rate by twelve and multiply the five years by twelve. Neither move is optional.
\[ A = 2000\left(1 + \frac{0.06}{12}\right)^{12(5)} = 2000(1.005)^{60} \]
Compute and compare
Why: The correct amount is 2,697.70 dollars. It sits above the annual answer, as it must, because paying interest sooner always earns a little more.
\[ A \approx 2697.70 \;\checkmark \qquad 2676.45 < 2697.70 \]
Notation
Annotate
From Trap: reading the word monthly and only half using it — read this one piece at a time. What is each part doing?
On: \( A \approx 2697.70 \;\checkmark \qquad 2676.45 < 2697.70 \)
Concept
Push the number of periods past monthly, past daily, past every second - to the limit. That is continuous compounding, and it is where e shows up in the formula.
\[ A = P\,e^{rt} \]
Notice what disappeared: there is no letter for periods per year, because there is no longer a finite number of them.
| formula | use it when | letters |
|---|---|---|
| A = P(1 + r/n) raised to nt | the problem names a compounding count | P, r, n, t |
| A = P times e to the rt | the problem says continuously | P, r, t |
Continuous compounding always gives the largest amount for a given rate and time - but only barely more than daily. It is the ceiling, not a jackpot.
Estimation
Predict first
An account holds 7,500 dollars at an annual rate of 4.5 percent, compounded continuously. Find the balance after 8 years.
Commit before you compute: what does Worked example: compounded continuously come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify against monthly compounding on the same account
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Monthly compounding of the same deposit gives 10,742.73 dollars.
Worked example
An account holds 7,500 dollars at an annual rate of 4.5 percent, compounded continuously. Find the balance after 8 years.
Choose the formula from the wording
Why: The word continuously rules out the periodic formula. There is no number of periods to substitute, so the natural base formula is the only option.
\[ A = P\,e^{rt}, \qquad P = 7500, \quad r = 0.045, \quad t = 8 \]
Multiply the rate by the time first
Why: The exponent is a single product. Doing it before touching the exponential key prevents the classic calculator error of raising to only part of the exponent.
\[ rt = (0.045)(8) = 0.36 \]
Evaluate the growth factor
Why: Raising e to the 0.36 power gives about 1.4333, so the account grows by roughly 43 percent over the eight years.
\[ e^{0.36} = 1.433329415\ldots \]
Multiply by the principal and round to cents
Why: Rounding is saved for the final line so no precision is lost in the middle of the calculation.
\[ A = 7500(1.433329415) = 10749.97\ldots \;\Rightarrow\; A \approx 10749.97 \text{ dollars} \]
Verify against monthly compounding on the same account
Why: Monthly compounding of the same deposit gives 10,742.73 dollars. Continuous must be slightly larger and it is, by about seven dollars - a believable gap, not a suspicious one.
\[ 7500\left(1 + \tfrac{0.045}{12}\right)^{96} = 10742.73 \;<\; 10749.97 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "compounded continuously", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Monthly compounding of the same deposit gives 10,742.73 dollars. Continuous must be slightly larger and it is, by about seven dollars - a believable gap, not a suspicious one.
Concept
Same deposit of 2,000 dollars, same 6 percent, same 5 years. Only the compounding frequency changes.
| compounding | n | balance in dollars |
|---|---|---|
| annually | 1 | 2676.45 |
| semiannually | 2 | 2687.83 |
| quarterly | 4 | 2693.71 |
| monthly | 12 | 2697.70 |
| daily | 365 | 2699.65 |
| continuously | no n | 2699.72 |
Annual to monthly is worth about 21 dollars. Monthly to continuous is worth about 2 dollars. Daily to continuous is worth seven cents.
This is the same convergence you saw in the table for e, dressed in money. The frequency matters, then it stops mattering - but the rate and the time never stop mattering.
Pattern
Step through it
Step through How much does compounding more often actually buy you? one row at a time. What is driving the change, and what would the row after the last one be?
Constraint
Discussion prompt
Run Pattern: any interest problem, start to finish with this step confiscated:
Otherwise divide the rate by the period count AND multiply the years by that same count - both, every time.
Is it still possible? If it is, say what takes its place and what it costs you. If it is not, say exactly what that step was providing that nothing else does.
Hint: A step you can drop for free was never load-bearing. If you cannot drop it, name the thing that goes wrong the moment it is gone.
Answer:
Pattern
One extra habit worth building: if the question asks for the interest earned rather than the balance, subtract the principal at the end.
\[ \text{interest} = A - P \]
Edge cases
Discussion prompt
Pattern: any interest problem, start to finish works on the cases you have just seen. Push it to the edge: what is the most degenerate input it still handles — empty, zero, one item, everything equal — and what is the first case where it stops being true? Name the case, not just "it breaks".
Hint: Try the smallest legal input, then the largest, then the one where two things collide. Methods are specified at their edges; the middle takes care of itself.
Answer:
One extra habit worth building: if the question asks for the interest earned rather than the balance, subtract the principal at the end.
Prediction
Predict first
How much is in the account after 3 years, to the nearest cent?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: 4643.02 dollars
Why: Quarterly means n = 4, so the per-period rate is 0.05/4 = 0.0125 and the number of periods is 4 times 3 = 12. Then 4000 times 1.0125 raised to the 12th power is 4000 times 1.16075451, which is 4643.02 dollars.
Check
Work it fully before choosing. Watch the two places the compounding count has to appear.
\[ P = 4000, \quad r = 0.05, \quad \text{quarterly}, \quad t = 3 \]
Check your understanding
How much is in the account after 3 years, to the nearest cent?
Answer: A
Why: Quarterly means n = 4, so the per-period rate is 0.05/4 = 0.0125 and the number of periods is 4 times 3 = 12. Then 4000 times 1.0125 raised to the 12th power is 4000 times 1.16075451, which is 4643.02 dollars.
Section
Part 5
Concept
Populations, bank balances, radioactive samples, drug concentrations, resale values - once you strip the story away, they are all the same object.
\[ A(t) = A_{0}\,b^{\,t} \]
initial amount — The value of the quantity when the clock reads zero. It is the number out front, and it is the output at time zero because any base to the zero power is one.
The base carries all of the behavior. Above one and the story is growth; between zero and one and the story is decay.
When the situation is continuous rather than step-by-step, the same model is written with the natural base and a rate constant in the exponent.
\[ A(t) = A_{0}\,e^{\,kt} \qquad k > 0 \text{ grows}, \quad k < 0 \text{ decays} \]
The units of time have to match whatever the base or the rate constant was measured in. Years with years, hours with hours - always.
Picture it
Animation
Shows: Reading an exponential model — a rendered Manim animation.
Rendered with Manim.
Takeaway: A factor above one grows; below one decays. That is the whole reading.
Concept
Word problems almost never hand you the base. They hand you a percent, and you convert.
Growing by a percent means you keep everything you had and add some, so the base is one plus the decimal rate.
\[ \text{grows } 7\% \text{ per year} \;\Longrightarrow\; b = 1 + 0.07 = 1.07 \]
Decaying by a percent means you keep what is left over, so the base is one minus the decimal rate.
\[ \text{loses } 15\% \text{ per year} \;\Longrightarrow\; b = 1 - 0.15 = 0.85 \]
| the words | decimal rate | base to use |
|---|---|---|
| grows 3 percent a year | 0.03 | 1.03 |
| grows 100 percent a year (doubles) | 1.00 | 2 |
| loses 8 percent a year | 0.08 | 0.92 |
| loses half each year | 0.50 | 0.5 |
The single most common slip here is using the percent itself as the base. A base of 0.15 would mean losing 85 percent a year, not 15.
Comparison
Comparison matrix
From Turning a percent into a base: refill the base to use column from what you know. The rest of the table is as it appeared.
| the words | decimal rate | base to use |
|---|---|---|
| grows 3 percent a year | 0.03 | 1.03 |
| grows 100 percent a year (doubles) | 1.00 | 2 |
| loses 8 percent a year | 0.08 | 0.92 |
| loses half each year | 0.50 | 0.5 |
Ranking
Put in order
Put the moves of Worked example: a car losing value into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The purchase price is the value at time zero, so it is the number that sits out front of the model.
Worked example
A car is bought for 24,000 dollars and loses 15 percent of its value every year. Write the model and find the value after 4 years.
Identify the initial amount
Why: The purchase price is the value at time zero, so it is the number that sits out front of the model.
\[ A_{0} = 24000 \]
Convert the 15 percent loss into a base
Why: Losing 15 percent means keeping 85 percent, so each year the value is multiplied by 0.85. That fraction is the base, not the 0.15.
\[ b = 1 - 0.15 = 0.85 \;\Longrightarrow\; V(t) = 24000(0.85)^{t} \]
Substitute four years and evaluate the power
Why: The exponent counts years because the 15 percent was quoted per year. Raising 0.85 to the fourth power gives the fraction of the original value that survives.
\[ V(4) = 24000(0.85)^{4} = 24000(0.52200625) \]
Multiply and interpret
Why: About 52 percent of the value remains, so the car is worth a little over half what it cost.
\[ V(4) = 12528.15 \text{ dollars} \]
Verify by stepping down one year at a time
Why: Multiplying by 0.85 four times in a row must land on the same number as raising 0.85 to the fourth power, and it does - exactly, with no rounding.
| year | value in dollars |
|---|---|
| 0 | 24000 |
| 1 | 20400 |
| 2 | 17340 |
| 3 | 14739 |
| 4 | 12528.15 |
Picture it
Animation
Shows: Each line of the worked example "a car losing value", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Multiplying by 0.85 four times in a row must land on the same number as raising 0.85 to the fourth power, and it does - exactly, with no rounding.
Concept
Two points is exactly enough to pin down an exponential, because the model has exactly two unknowns: the number out front and the base.
\[ y = a\,b^{x} \]
If one of the points sits at input zero, you get the number out front for free, and only the base is left to find.
\[ (0, a) \text{ given} \;\Longrightarrow\; a \text{ is known immediately} \]
Then substitute the second point, divide away the known coefficient, and take the appropriate root to release the base.
\[ a\,b^{\,k} = y_{1} \;\Longrightarrow\; b^{\,k} = \frac{y_{1}}{a} \;\Longrightarrow\; b = \sqrt[k]{\frac{y_{1}}{a}} \]
If neither point sits at zero, divide one equation by the other. The coefficient cancels and you are left with a single power of the base.
Hypothesis
Predict first
Worked example: a town's population is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Set the clock so that one point lands at zero
Why: Letting time be measured in years since 2015 makes the first data point the initial amount, which removes one unknown before any algebra happens.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
A town had 5,000 residents in 2015 and 8,640 residents in 2018. Assume exponential growth. Find the model and predict the population in 2021.
Set the clock so that one point lands at zero
Why: Letting time be measured in years since 2015 makes the first data point the initial amount, which removes one unknown before any algebra happens.
\[ t = \text{years since } 2015 \;\Longrightarrow\; A_{0} = 5000, \quad P(3) = 8640 \]
Substitute the second point and isolate the power
Why: Dividing both sides by 5,000 strips off the known coefficient and leaves the base raised to the third power alone.
\[ 5000\,b^{3} = 8640 \;\Longrightarrow\; b^{3} = \frac{8640}{5000} = 1.728 \]
Take the cube root to release the base
Why: The exponent is three, so the matching inverse move is a cube root. It comes out clean because 1.2 cubed is exactly 1.728.
\[ b = \sqrt[3]{1.728} = 1.2 \]
Write the model and say what the base means
Why: A base of 1.2 is one plus 0.20, so the town grows by 20 percent per year.
\[ P(t) = 5000(1.2)^{t} \]
Predict 2021 by using six years
Why: 2021 is six years after 2015, and population must be reported as a whole number of people.
\[ P(6) = 5000(1.2)^{6} = 5000(2.985984) = 14929.92 \approx 14930 \text{ people} \]
Verify the model reproduces the 2018 data point
Why: Substituting three years must return the given 8,640 exactly, since that point was used to build the model. It does, so no arithmetic slipped in.
\[ P(3) = 5000(1.2)^{3} = 5000(1.728) = 8640 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "a town's population", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Substituting three years must return the given 8,640 exactly, since that point was used to build the model. It does, so no arithmetic slipped in.
Step zero
Discussion prompt
Worked example: two points, neither at zero — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Write one equation per point
Answer:
Worked example
Find the exponential function through these two points.
\[ (1, 12) \quad \text{and} \quad (3, 108) \]
Write one equation per point
Why: Substituting each point into the general form gives a system in the two unknowns, the coefficient and the base.
\[ a\,b^{1} = 12 \qquad\text{and}\qquad a\,b^{3} = 108 \]
Divide the second equation by the first
Why: Division is the move that kills the unknown coefficient: it cancels top and bottom, and the powers of the base subtract.
\[ \frac{a\,b^{3}}{a\,b^{1}} = \frac{108}{12} \;\Longrightarrow\; b^{2} = 9 \]
Take the positive square root only
Why: Negative three is also a square root of nine, but a base must be positive, so it is rejected on definition grounds - not by accident.
\[ b = 3 \quad (b = -3 \text{ rejected since } b > 0) \]
Back-substitute to find the coefficient
Why: Use the simpler of the two original equations. With the base known, the coefficient falls out in one division.
\[ a(3) = 12 \;\Longrightarrow\; a = 4 \;\Longrightarrow\; y = 4\cdot 3^{x} \]
Verify both original points
Why: A model built from two points must reproduce both of them. Substituting one and three returns 12 and 108 exactly, so the function is confirmed.
\[ 4\cdot 3^{1} = 12 \;\checkmark \qquad 4\cdot 3^{3} = 4(27) = 108 \;\checkmark \]
Concept
Some quantities are described by how long they take to fall to half of whatever they currently are - medicine in your blood, carbon in a fossil, a radioactive sample.
half-life — The amount of time it takes a decaying quantity to drop to one half of its current value. It is the same length of time no matter when you start the clock.
That last sentence is the surprising part. Going from 200 to 100 takes exactly as long as going from 50 to 25.
So the natural model uses a base of one half, with the exponent counting how many half-lives have gone by.
\[ A(t) = A_{0}\left(\tfrac{1}{2}\right)^{t/h} \qquad h = \text{the half-life} \]
Dividing the elapsed time by the half-life is the whole trick. If that quotient is a whole number, you can finish the problem by halving repeatedly and never touch a calculator.
Picture it
Animation
Shows: Half-life is constant — a rendered Manim animation.
Rendered with Manim.
Takeaway: Every equal time step multiplies by the same factor, wherever you start.
Missing information
Discussion prompt
Caffeine has a half-life of about 5 hours. You drink a coffee containing 200 milligrams. How much is left 15 hours later?
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
The dose is the amount at time zero, and the half-life is the number the elapsed time gets divided by.
Worked example
Caffeine has a half-life of about 5 hours. You drink a coffee containing 200 milligrams. How much is left 15 hours later?
Name the initial amount and the half-life
Why: The dose is the amount at time zero, and the half-life is the number the elapsed time gets divided by.
\[ A_{0} = 200, \quad h = 5, \quad t = 15 \]
Write the model
Why: The base is one half because the quantity halves, and the exponent measures elapsed time in units of half-lives.
\[ A(t) = 200\left(\tfrac{1}{2}\right)^{t/5} \]
Convert 15 hours into a number of half-lives
Why: Fifteen hours divided by a five-hour half-life is three, so three complete halvings have happened.
\[ \frac{t}{h} = \frac{15}{5} = 3 \]
Evaluate the power and multiply
Why: One half cubed is one eighth, and one eighth of the 200 milligram dose is 25 milligrams.
\[ A(15) = 200\left(\tfrac{1}{2}\right)^{3} = 200\cdot\tfrac{1}{8} = 25 \text{ mg} \]
Verify by halving three times by hand
Why: Each five-hour block cuts the amount in half. Three blocks take 200 to 100 to 50 to 25, matching the formula with no rounding anywhere.
| hours elapsed | half-lives | caffeine in mg |
|---|---|---|
| 0 | 0 | 200 |
| 5 | 1 | 100 |
| 10 | 2 | 50 |
| 15 | 3 | 25 |
Picture it
Animation
Shows: Each line of the worked example "caffeine in your bloodstream", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Each five-hour block cuts the amount in half. Three blocks take 200 to 100 to 50 to 25, matching the formula with no rounding anywhere.
Concept
A hot drink in a cool room does not cool toward zero. It cools toward room temperature, and it slows down as it gets close.
That is a decaying exponential sitting on top of a shifted asymptote - exactly the transformation you learned three parts ago, now doing real work.
\[ T(t) = T_{s} + (T_{0} - T_{s})e^{-kt} \]
| symbol | meaning |
|---|---|
| T of t | temperature at time t |
| T sub s | surrounding (room) temperature - the asymptote |
| T sub 0 | starting temperature of the object |
| k | a positive cooling constant for that object |
The gap between the object and the room is what decays. When the exponent drives that term to nearly zero, the temperature is nearly the room's.
\[ \text{As } t \to \infty, \quad e^{-kt} \to 0 \quad\Longrightarrow\quad T(t) \to T_{s} \]
Trade off
Comparison matrix
From Newton's law of cooling: every row here is a choice with a cost. Fill the meaning column, then say which row you would actually pick and what you give up for it.
| symbol | meaning |
|---|---|
| T of t | temperature at time t |
| T sub s | surrounding (room) temperature - the asymptote |
| T sub 0 | starting temperature of the object |
| k | a positive cooling constant for that object |
Estimation
Predict first
Coffee is poured at 180 degrees Fahrenheit into a room held at 70 degrees. Its cooling constant is 0.07 per minute. Find the temperature after 10 minutes.
Commit before you compute: what does Worked example: how fast the coffee cools come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify with the two anchors the model must satisfy
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. At time zero the model must return the pouring temperature, and every later value must sit strictly between the room temperature and the starting temperature.
Worked example
Coffee is poured at 180 degrees Fahrenheit into a room held at 70 degrees. Its cooling constant is 0.07 per minute. Find the temperature after 10 minutes.
Identify the room temperature and the starting gap
Why: The room temperature is the asymptote the drink is falling toward, and the gap is what actually decays.
\[ T_{s} = 70, \quad T_{0} = 180 \;\Longrightarrow\; T_{0} - T_{s} = 110 \]
Write the model for this cup
Why: Substituting the room temperature, the gap, and the cooling constant turns the general law into a specific function of time in minutes.
\[ T(t) = 70 + 110\,e^{-0.07t} \]
Substitute ten minutes and compute the exponent
Why: The exponent is negative because the gap is shrinking, and it must be evaluated as a single product before the exponential is applied.
\[ -0.07(10) = -0.7 \;\Longrightarrow\; e^{-0.7} = 0.4965853\ldots \]
Finish the arithmetic
Why: The gap has shrunk from 110 degrees to about 54.6 degrees, and adding the room temperature back gives the actual reading.
\[ T(10) = 70 + 110(0.4965853) = 70 + 54.62 = 124.62 \text{ degrees} \]
Verify with the two anchors the model must satisfy
Why: At time zero the model must return the pouring temperature, and every later value must sit strictly between the room temperature and the starting temperature. Both hold here.
\[ T(0) = 70 + 110(1) = 180 \;\checkmark \qquad 70 < 124.62 < 180 \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "how fast the coffee cools", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At time zero the model must return the pouring temperature, and every later value must sit strictly between the room temperature and the starting temperature. Both hold here.
Elimination
Eliminate the wrong options
How much of the sample remains after 32 days?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: 32 days divided by the 8-day half-life is 4 half-lives, so the amount is halved four times: 240, then 120, then 60, then 30, then 15. In formula form, 240 times one half to the fourth power is 240 divided by 16, which is 15 mg.
Check
Count the half-lives first. If the count is a whole number you will not need a calculator at all.
A 240 milligram sample of an isotope has a half-life of 8 days.
Check your understanding
How much of the sample remains after 32 days?
Answer: A
Why: 32 days divided by the 8-day half-life is 4 half-lives, so the amount is halved four times: 240, then 120, then 60, then 30, then 15. In formula form, 240 times one half to the fourth power is 240 divided by 16, which is 15 mg.
Section
Part 6
Concept
An exponential function never repeats an output. Different inputs always give different outputs, because the curve only ever climbs or only ever falls.
So if two powers of the same base are equal, the exponents themselves have to be equal. There is no other way it could happen.
\[ b^{\,m} = b^{\,n} \quad\Longleftrightarrow\quad m = n \qquad (b > 0,\; b \ne 1) \]
equal-bases property — If both sides of an equation are written as powers of one and the same base, you may set the exponents equal to each other and drop the bases.
Read the condition carefully. It says the same base. The property says nothing at all about two powers of different bases.
Explain it
Discussion prompt
Explain The equal-bases property to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
An exponential function never repeats an output. Different inputs always give different outputs, because the curve only ever climbs or only ever falls.
Intuition
Think of powers of two as a list of labeled boxes: 1, 2, 4, 8, 16, 32, and so on, with the exponent printed on the side of each box.
No two boxes hold the same number. Sixteen shows up once and only once, on the box labeled four.
So if someone tells you two of these boxes contain equal amounts, they are describing the same box - which means the labels match. That is all the property says.
And it fails the moment the boxes come from different shelves. Eight appears on the two-shelf labeled three and on the eight-shelf labeled one; equal contents, wildly different labels.
\[ 2^{3} = 8^{1} = 8 \quad\text{but}\quad 3 \ne 1 \]
Analogy
Discussion prompt
Explain Why the property is honest by analogy to something with no College Algebra in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Think of powers of two as a list of labeled boxes: 1, 2, 4, 8, 16, 32, and so on, with the exponent printed on the side of each box.
Step zero
Discussion prompt
Worked example: rewrite both sides in base two — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Find the largest base both sides are powers of
Answer:
Worked example
Solve for the variable.
\[ 8^{\,x+1} = 32^{\,x-1} \]
Find the largest base both sides are powers of
Why: Eight and thirty-two are both powers of two. Choosing two makes both sides expressible without any fractions or roots.
\[ 8 = 2^{3}, \qquad 32 = 2^{5} \]
Rewrite each side and use the power-of-a-power rule
Why: A power raised to a power multiplies the exponents. The exponent in parentheses must be multiplied as a whole, so keep the parentheses until you distribute.
\[ \left(2^{3}\right)^{x+1} = \left(2^{5}\right)^{x-1} \;\Longrightarrow\; 2^{\,3(x+1)} = 2^{\,5(x-1)} \]
Drop the matching bases and set the exponents equal
Why: Both sides are now powers of the same base, so the equal-bases property applies and the problem becomes ordinary linear algebra.
\[ 3(x+1) = 5(x-1) \]
Distribute across every term and solve
Why: The outer factor multiplies both the variable and the constant inside the parentheses. Collect the variable on one side and the numbers on the other.
\[ 3x + 3 = 5x - 5 \;\Longrightarrow\; 8 = 2x \;\Longrightarrow\; x = 4 \]
Verify in the original equation
Why: Substituting four gives eight to the fifth on the left and thirty-two cubed on the right. Both equal 32,768, so the two sides genuinely agree.
\[ 8^{5} = 32768 \qquad 32^{3} = 32768 \qquad \checkmark \]
Worked example
The right side is a fraction, which is a signal that a negative exponent is coming.
\[ 2^{\,3x} = \frac{1}{16} \]
Write the denominator as a power of two
Why: Sixteen is two to the fourth, so the fraction is one over two to the fourth. That is the only rewriting the right side needs.
\[ \frac{1}{16} = \frac{1}{2^{4}} \]
Convert the reciprocal into a negative exponent
Why: One over a power equals that base to the negative exponent. This is the rule that lets a fraction join the same base as the left side.
\[ \frac{1}{2^{4}} = 2^{-4} \;\Longrightarrow\; 2^{\,3x} = 2^{-4} \]
Set the exponents equal and solve
Why: Same base on both sides, so the exponents must match. Dividing by three leaves a fraction, which is a perfectly acceptable answer.
\[ 3x = -4 \;\Longrightarrow\; x = -\frac{4}{3} \]
Verify by substituting the fraction back in
Why: Three times negative four thirds is negative four, and two to the negative fourth is one sixteenth - exactly the right side of the original equation.
\[ 2^{\,3\left(-\frac{4}{3}\right)} = 2^{-4} = \frac{1}{16} \;\checkmark \]
Picture it
Animation
Shows: Doubling time — a rendered Manim animation.
Rendered with Manim.
Takeaway: Independent of where you start — a defining property of exponential growth.
Ranking
Put in order
Put the moves of Worked example: a radical on one side into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Four is two squared, so the left side becomes two raised to twice the exponent.
Worked example
Radicals are fractional exponents in disguise. Convert first, then everything looks familiar.
\[ 4^{\,x+1} = \sqrt{8} \]
Rewrite the left side in base two
Why: Four is two squared, so the left side becomes two raised to twice the exponent.
\[ 4^{\,x+1} = \left(2^{2}\right)^{x+1} = 2^{\,2x+2} \]
Turn the square root into a one-half power
Why: A square root is the one-half power, and eight is two cubed, so the right side is two raised to three halves.
\[ \sqrt{8} = 8^{1/2} = \left(2^{3}\right)^{1/2} = 2^{3/2} \]
Set the exponents equal
Why: Both sides are powers of two now, so the exponents must agree even though one of them is a fraction.
\[ 2x + 2 = \frac{3}{2} \]
Solve the linear equation with the fraction
Why: Subtract two from both sides, then divide by two. Writing two as four halves keeps the arithmetic in one common denominator.
\[ 2x = \frac{3}{2} - \frac{4}{2} = -\frac{1}{2} \;\Longrightarrow\; x = -\frac{1}{4} \]
Verify in the original equation
Why: The exponent becomes three quarters, and four to the three quarters is two to the three halves, which is two times the square root of two - exactly the square root of eight.
\[ 4^{\,-\frac{1}{4}+1} = 4^{3/4} = 2^{3/2} = 2\sqrt{2} = \sqrt{8} \;\checkmark \]
Picture it
Animation
Shows: Each line of the worked example "a radical on one side", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The exponent becomes three quarters, and four to the three quarters is two to the three halves, which is two times the square root of two - exactly the square root of eight.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The mistake: the equal-bases property gets remembered as drop the bases, and the word equal quietly falls off.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The exponents are set equal without ever converting.
Convert to a genuine common base before touching the exponents, then distribute across every term.
Why: The exponents are set equal without ever converting. The resulting statement is impossible, so the student reports no solution.
Trap
The mistake: the equal-bases property gets remembered as drop the bases, and the word equal quietly falls off.
\[ 2^{\,x+1} = 8^{\,x-3} \]
Version one: cross off a 2 and an 8 as if they matched
Why: The exponents are set equal without ever converting. The resulting statement is impossible, so the student reports no solution.
\[ x + 1 \;\overset{?}{=}\; x - 3 \;\Longrightarrow\; 1 = -3 \;\text{(no solution - wrong)} \]
Version two: convert, then distribute only halfway
Why: The three multiplies the variable but not the constant, so the equation solved is not the one the conversion produced.
\[ x + 1 \;\overset{?}{=}\; 3x - 3 \;\Longrightarrow\; x = 2 \]
Test that answer and watch it fail
Why: Substituting two gives eight on the left and one eighth on the right. Those are not equal, so two is not a solution.
\[ 2^{3} = 8 \quad\text{but}\quad 8^{-1} = \tfrac{1}{8} \quad \times \]
Convert to a genuine common base before touching the exponents, then distribute across every term.
\[ 2^{\,x+1} = 8^{\,x-3}, \qquad 8 = 2^{3} \]
Rewrite the right side in base two
Why: Eight is two cubed, and a power of a power multiplies exponents, keeping the whole exponent inside parentheses.
\[ 2^{\,x+1} = \left(2^{3}\right)^{x-3} = 2^{\,3(x-3)} \]
Now the bases match, so set the exponents equal and distribute fully
Why: The three multiplies both the x and the negative three inside the parentheses. Missing that constant is what produced the wrong answer.
\[ x + 1 = 3(x - 3) = 3x - 9 \;\Longrightarrow\; 10 = 2x \;\Longrightarrow\; x = 5 \]
Verify in the original
Why: Substituting five gives two to the sixth on the left and eight squared on the right. Both are 64, so the solution is genuine - and it existed all along.
\[ 2^{6} = 64 \qquad 8^{2} = 64 \qquad \checkmark \]
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Ranking
Put in order
These are the steps of Pattern: solving with a common base, scrambled. Put them back in order before the next slide shows you.
Why: This is the order the recipe itself gives. Recalling the sequence without the slide in front of you is the difference between recognising the method and being able to run it — most of what goes wrong in practice is a step done out of turn.
Pattern
Useful conversions to have memorized cold, because they turn up constantly.
| expression | as a power of the base |
|---|---|
| 1 | base to the 0 |
| the base itself | base to the 1 |
| 1 over the base to the k | base to the negative k |
| square root of the base | base to the one half |
| cube root of the base | base to the one third |
If step 2 is impossible - the two sides are simply not powers of a common number - the common-base method is the wrong tool. That is where logarithms come in.
Comparison
Comparison matrix
From Pattern: solving with a common base: refill the as a power of the base column from what you know. The rest of the table is as it appeared.
| expression | as a power of the base |
|---|---|
| 1 | base to the 0 |
| the base itself | base to the 1 |
| 1 over the base to the k | base to the negative k |
| square root of the base | base to the one half |
| cube root of the base | base to the one third |
Check
Both numbers are powers of the same base. Convert first, then distribute carefully.
\[ 25^{\,x+1} = 125^{\,x-1} \]
Check your understanding
Solve for x.
Answer: A
Why: Write 25 as 5 squared and 125 as 5 cubed, giving exponents 2(x+1) and 3(x-1). Setting them equal: 2x + 2 = 3x - 3, so x = 5. Check: 25 to the 6th and 125 to the 4th are both 5 to the 12th.
Pattern
Six parts, one decision map. Read the question, find your row, use that tool.
| what you are handed | the move |
|---|---|
| a table of values | differences for linear, ratios for exponential |
| an equation to graph | base sets growth or decay; outside constant sets the asymptote |
| a percent per period | base is one plus or one minus the decimal rate |
| a deposit and a compounding word | periodic formula, or the natural base formula if continuous |
| a half-life | base one half, exponent is time divided by the half-life |
| two data points | divide the equations to cancel the coefficient, then root the base out |
| the variable in an exponent | common base if possible, logarithms otherwise |
Notice how many rows are really the same idea wearing a costume. Initial amount, constant multiplier, exponent that counts steps - that is every row above.
Real world
Discussion prompt
Outside this lesson: where does Exponential Functions and Growth actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of Pattern: which tool does this problem want? is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
That deck builds exponential functions from the ground up. It contrasts a constant rate with a constant ratio in a table, explains the restrictions on the base and why they exist, and covers growth and decay graphs with their horizontal asymptote and the transformations that move it. It then introduces the natural base e as the limit of compounding more and more often, computes compound and continuous interest with real numbers, and works through growth, decay, half-life, and cooling models, fitting a model to two data points, and solving exponential equations with a common base. It targets four killer errors: confusing a variable base with a variable exponent, multiplying the base by the exponent, forgetting that a vertical shift moves the horizontal asymptote, and using the annual interest formula when the problem says monthly.
Concept
Everything in Part 6 depended on one lucky circumstance: both sides happened to be powers of the same number.
\[ 2^{x} = 7 \]
Seven is not a power of two. There is no rewriting that will make it one, so there is nothing to match.
But a solution obviously exists: the doubling curve passes through every positive height exactly once, so it passes through seven somewhere between two and three.
\[ 2^{2} = 4 < 7 < 8 = 2^{3} \;\Longrightarrow\; 2 < x < 3 \]
To name that number exactly, you need a tool that pulls a variable out of an exponent - the inverse of the exponential function. That tool is the logarithm, and it is the next deck.
Counterexample
Discussion prompt
Everything in Part 6 depended on one lucky circumstance: both sides happened to be powers of the same number.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
Seven is not a power of two. There is no rewriting that will make it one, so there is nothing to match.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Constant Rate or Constant Ratio · The Exponential Function · Transformations and the Moving Asymptote · The Natural Base e and Compounding · Growth, Decay, and Real Models · Solving with a Common Base. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
One sentence holds this whole topic together: exponential means multiply by the same number each step, and everything else follows from that.
| situation | what to write |
|---|---|
| constant ratio in a table | initial value times ratio to the x |
| grows p percent per period | base one plus p as a decimal |
| loses p percent per period | base one minus p as a decimal |
| compounded n times a year | rate over n, exponent n times t |
| compounded continuously | natural base, exponent r times t |
| half-life h | base one half, exponent t over h |
| same base on both sides | set the exponents equal |
And the four errors worth taping to your notebook: variable in the wrong slot, base multiplied by the exponent, asymptote left behind after a shift, and the compounding count used in only one of its two places.
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