The form and graph of an exponential function, growth against decay, compound interest and where the number e comes from, the logarithm as the exponential's inverse, the three laws of logarithms and why they are exponent laws read backwards, change of base, solving exponential and logarithmic equations, and the hyperbolic functions.
Subject: Calculus I · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Calculus I · Chapter 1 — Functions and Graphs
Exponential and Logarithmic Functions
Objectives
Five outcomes. The second explains why one particular base, e, is the one calculus uses for everything.
OpenStax Calculus Volume 1, §1.5 Exponential and Logarithmic Functions §1.5, pp. 80-98 — the section these objectives are drawn from
Warm-up
Section 1.4 showed how an inverse is built. This section's two families are the most important example of the pattern.
Discussion prompt
The equation 2 to the power x equals 8 has the answer 3. But what is the answer to 2 to the power x equals 10? Can you write it down at all with the tools of Chapter 1 so far?
Hint: The answer exists — the exponential is continuous and increasing — but no polynomial or root expresses it.
Answer:
The answer is between 3 and 4, since 2 cubed is 8 and 2 to the fourth is 16. It exists and is unique, because the exponential is strictly increasing and therefore one-to-one.
\[ 2^x = 10 \;\Longrightarrow\; x = \log_2 10 \approx 3.3219 \]
But no combination of arithmetic and roots produces it. The number needs a name, and the logarithm is that name. It is the inverse function of Section 1.4 applied to the exponential — which is exactly why it exists and exactly why it is unique.
Concept
An exponential function multiplies by a fixed factor each time the input increases by one. That makes it strictly monotone, hence one-to-one, hence invertible — and its inverse is called the logarithm.
exponential function — A function of the form f of x equals b to the power x, where the base b is positive and not equal to one. Its domain is all real numbers and its range is the positive numbers.
\[ f(x) = b^x, \quad b > 0, \; b \ne 1 \]
Contrast this with a linear function, which ADDS a fixed amount per unit of input. Adding gives a straight line; multiplying gives a curve that eventually outgrows every polynomial, however high its degree.
Figure (svg): Exponential growth and exponential decay on one pair of axes, both passing through the same y-intercept
OpenStax Calculus Volume 1, §1.5 Exponential and Logarithmic Functions §1.5, pp. 80-83
Section
Section 1
Concept
For a base larger than one the outputs grow as the input grows; for a base between zero and one they shrink. Every exponential is positive everywhere, passes through the point zero comma one, and approaches the horizontal axis on one side without ever reaching it.
growth and decay — An exponential with base greater than one is increasing and models growth; with base between zero and one it is decreasing and models decay. Since one over b to the x equals b to the negative x, a decay function is a growth function with the input reflected.
\[ b > 1: \text{ growth}; \qquad 0 < b < 1: \text{ decay} \]
The output at input zero is always one, because any non-zero base to the zero power is one. That fixed point is why the coefficient in front, in a model like A times b to the x, is exactly the initial amount.
Figure (svg): Exponential growth and exponential decay on one pair of axes, both passing through the same y-intercept
OpenStax Calculus Volume 1, §1.5 Exponential and Logarithmic Functions §1.5, pp. 80-84 — exponential functions and their graphs
Picture it
Base 2 and base one half, on the same axes.
Figure (svg): Exponential growth and exponential decay on one pair of axes, both passing through the same y-intercept
The two curves are mirror images in the vertical axis, because one half to the x equals 2 to the negative x. The horizontal axis is an asymptote for both, on opposite sides — which is the picture behind the limits at infinity of Section 4.6.
Worked example
Example 1.37. The constant multiplier is the whole model.
\[ \text{A colony of 1000 bacteria doubles every hour. Find the population after } t \text{ hours, and at } t = 5. \]
Identify the initial amount
Why: The output when the input is zero.
\[ A = 1000 \]
Identify the multiplier per unit of input
Why: Doubling means multiplying by 2 each hour.
\[ b = 2 \]
Write the model
Why: Initial amount times base to the input.
\[ P(t) = 1000 \cdot 2 ^{t} \]
Evaluate at 5 hours
Why: Two to the fifth is 32.
\[ P(5) = 32000 \]
Figure (svg): The solution to Worked example a population model shown as a ladder of expressions, one row per legal move
\[ P(t) = 1000 \cdot 2^{t}, \qquad P(5) = 32{,}000 \]
Verify: double five times by hand
Why: Starting at 1000: after one hour 2000, then 4000, 8000, 16000, and 32000 after five. The step-by-step doubling agrees with the formula. Note how fast this is compared with linear growth — adding 1000 per hour would have reached only 6000. That divergence is the characteristic behaviour of exponentials and the reason they eventually beat every polynomial.
OpenStax Calculus Volume 1, §1.5 Exponential and Logarithmic Functions §1.5, pp. 82-83
Sorting
Look only at the base.
Sort into buckets
Sort each function.
The third and fifth are the ones to read carefully: a negative sign in the exponent flips growth into decay, because 2 to the negative x is the same as one half to the x. Always ask what the EFFECTIVE base is once the sign is absorbed.
Worked example
Checkpoint 1.37. A base below one, written two ways.
\[ \text{A drug's concentration falls by } 30\% \text{ each hour from } 80 \text{ mg. Model it.} \]
Convert the percentage loss to a multiplier
Why: Losing 30 percent leaves 70 percent.
\[ b = 0.7 \]
Write the model
Why: Initial amount times the multiplier to the input.
\[ C(t) = 80(0.7) ^{t} \]
Evaluate after 3 hours
Why: 0.7 cubed is 0.343.
\[ C(3) = 27.44 m g \]
Note the equivalent negative-exponent form
Why: The base below one can be written as a reciprocal.
\[ C(t) = 80(\frac{10}{7}) ^{-t} \]
Figure (svg): The solution to Worked example exponential decay shown as a ladder of expressions, one row per legal move
\[ C(t) = 80(0.7)^{t}, \qquad C(3) \approx 27.44 \text{ mg} \]
Verify: step down hour by hour
Why: From 80: after one hour 56, then 39.2, then 27.44. The formula agrees. The commonest modelling error is using 0.3 as the base rather than 0.7 — that would model keeping 30 percent, not losing it, and would give only 2.16 mg after three hours. Reading the percentage as what REMAINS is the check that catches it.
OpenStax Calculus Volume 1, §1.5 Exponential and Logarithmic Functions §1.5, pp. 83-84
Trap
\[ \text{falls } 30\% \text{ per hour from } 80 \text{ mg} \]
Take the base to be the percentage named
Why: The student reads 30 percent and writes 0.3.
\[ C(t) = 80(0.3)^t \;\Longrightarrow\; C(1) = 24 \quad \text{(wrong)} \]
Falling by 30 percent from 80 should leave 56, not 24. The model has thrown away 70 percent each hour instead of 30.
\[ C(t) = 80(0.7)^t \;\Longrightarrow\; C(1) = 56 \]
The base is what REMAINS, not what is lost
Why: Losing 30 percent means keeping 70 percent, so the multiplier is 1 minus 0.30.
The same rule the other way: growing by 30 percent means multiplying by 1.3, not by 0.3. Writing the base as one plus or minus the rate makes the direction explicit and removes the ambiguity entirely.
\[ b = 1 + r \text{ for growth}, \qquad b = 1 - r \text{ for decay} \]
Fill the middle
A quantity that loses 30 percent of itself each hour.
Fill in the blanks
b = 1 - 0.30 = 0.7
Why: Losing 30 percent leaves 70 percent, so the multiplier is 0.7. Writing the base as one minus the rate makes this automatic and prevents the classic error of using the rate itself as the base.
Prediction
Commit before reasoning.
Predict first
For f(x) = 2^-x, is there an input at which the output is exactly 0?
Correct: No. The outputs approach zero without ever attaining it.
\[ 2^{-x} > 0 \text{ for every real } x, \quad \text{but } 2^{-x} \to 0 \text{ as } x \to \infty \]
Why: A positive base raised to any real power is positive, so the output is never zero or negative. It can be made as small as you like by taking the input large enough, which is exactly what a horizontal asymptote means. This is the distinction between a limit and a value, and Section 2.2 will make it precise: the limit at infinity is 0 even though the function equals 0 nowhere. Practically, it is why a decay model never predicts a substance being completely gone.
Ranking
At x equal to 20, smallest first.
Put in order
Why: At x equal to 20 these are about 0.000001, then 20, then 400, then 3.2 million, then about 1.05 million million. The headline is the last one: the exponential has overtaken a fifth-degree polynomial and is running away. Exponential growth eventually beats every polynomial, no matter how high the degree — a fact Section 4.8 will prove with L'Hopital's rule.
Section
Section 2
Concept
Invest one unit at one hundred percent for one year. Compounding once gives two; compounding monthly gives more; daily more still. The values increase but do not grow without bound — they close in on a single number, and that number is called e.
the number e — The limit of one plus one over n, all raised to the power n, as n grows without bound. Its value is about 2.71828, and it is irrational.
\[ e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{n} \approx 2.718281828 \]
Continuous compounding at rate r for time t multiplies by e to the power rt. That model is the reason e appears in every growth and decay application, and Section 6.8 is devoted to it.
Figure (svg): A table of compounding frequencies converging to e, showing where the number comes from
OpenStax Calculus Volume 1, §1.5 Exponential and Logarithmic Functions §1.5, pp. 84-87 — compound interest and the number e
Picture it
One dollar at one hundred percent, compounded ever more often.
Figure (svg): A table of compounding frequencies converging to e, showing where the number comes from
By the time compounding is hourly the value has stopped moving in the fourth decimal place. That convergence is what makes continuous compounding a sensible idea rather than an infinite one, and it is the first genuine limit in this course.
Worked example
Example 1.39. Discrete compounding first, then continuous.
\[ \text{Invest } \$1000 \text{ at } 5\% \text{ for } 10 \text{ years, compounded monthly and then continuously.} \]
Write the discrete compounding formula
Why: Rate divided by frequency, compounded that many times per year.
\[ A = P(1 + \frac{r}{n}) ^{n t} \]
Substitute for monthly compounding
Why: Twelve periods a year for ten years.
\[ 1000(1 + \frac{0.05}{12}) ^{120} \]
Evaluate
Why: The multiplier is about 1.6470.
\[ \text{about } \$ 1647.01 \]
Now use the continuous formula
Why: The limit as the frequency grows.
\[ A = P e ^{r t} = 1000 e ^{0.5} \]
Evaluate
Why: e to the one half is about 1.6487.
\[ \text{about } \$ 1648.72 \]
Figure (svg): The solution to Worked example compound interest shown as a ladder of expressions, one row per legal move
\[ A_{\text{monthly}} \approx \$1647.01, \qquad A_{\text{cont}} = 1000e^{0.5} \approx \$1648.72 \]
Verify: check that continuous is the larger, but only slightly
Why: Continuous compounding must give more than any finite frequency, since the sequence increases toward its limit — and it does, by 1 dollar 71 cents over ten years. That the gap is so small is the practical lesson: past monthly, compounding more often buys almost nothing. If your continuous answer ever came out SMALLER than the monthly one, the arithmetic is wrong, because the limit is an upper bound here.
OpenStax Calculus Volume 1, §1.5 Exponential and Logarithmic Functions §1.5, pp. 86-87
Matching
Discrete or continuous, and at what frequency.
Match the pairs
Why: The middle two are the pair to keep straight: the rate is always divided by the frequency and the exponent always multiplied by it, so the annual rate is spread across the year. The continuous formula is what the second becomes in the limit as the frequency grows without bound.
Worked example
Checkpoint 1.39. The same formula outside finance.
\[ \text{A culture grows continuously at } 8\% \text{ per hour from } 500. \text{ Find it after } 6 \text{ hours.} \]
Write the continuous model
Why: Initial amount times e to the rate times time.
\[ N(t) = 500 e ^{0.08 t} \]
Substitute the time
Why: Six hours.
\[ N(6) = 500 e ^{0.48} \]
Evaluate the exponential
Why: e to the 0.48 is about 1.6161.
\[ \text{about } 808 \]
State with sensible precision
Why: A count of organisms.
\[ \text{about } 808\text{ organisms} \]
Figure (svg): The solution to Worked example continuous growth shown as a ladder of expressions, one row per legal move
\[ N(6) = 500e^{0.48} \approx 808 \]
Verify: bracket it against simple estimates
Why: At a flat 8 percent per hour without compounding, six hours would add 48 percent, giving 740. Continuous compounding must give more, and 808 does. It should also be less than doubling, since 8 percent for 6 hours is well short of the roughly 8.7 hours a continuous 8 percent rate needs to double. Both bounds hold, so the answer is in the right range.
OpenStax Calculus Volume 1, §1.5 Exponential and Logarithmic Functions §1.5, pp. 87-87
Error analysis
A student computes ten years of monthly compounding at five percent.
Annotate
On: \( A = 1000\left(1 + 0.05\right)^{120} \approx \$348{,}911 \)
The check that catches this instantly is a rough estimate: five percent for ten years cannot possibly turn a thousand dollars into three hundred thousand. Any compound interest answer should be sanity-checked against simple interest, which here would give 1500.
Fill the middle
One thousand at five percent for ten years, compounded continuously.
Fill in the blanks
A = 1000e^0.5 = 1000e^___}
Why: The exponent is the rate multiplied by the time, which is 0.5. Note that only the product matters: five percent for ten years gives the same multiplier as ten percent for five years, which is a genuinely useful shortcut.
Two truths and a lie
All three are about e and compounding.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one, and the table in the visual refutes it directly. Going from annual to monthly — twelve times as often — raised the multiplier only from 2.00 to 2.61, and going from daily to hourly changed it in the fourth decimal place. The returns diminish sharply because the sequence is converging, which is the whole reason a continuous limit exists at all.
Prediction
Commit before reasoning.
Predict first
Compounding more often increases the total. Why does it not increase without bound?
Correct: Because raising the frequency also shrinks the per-period rate, and the two effects nearly cancel.
\[ \left(1 + \tfrac{1}{n}\right)^{n}: \quad \text{base } \to 1, \; \text{exponent } \to \infty \]
Why: Doubling the frequency doubles the exponent but halves the rate inside the bracket, so the base moves closer to 1 by as much as the exponent grows. The net gain shrinks with each doubling, and the values converge to about 2.71828 rather than escaping. This tension between a base approaching 1 and an exponent approaching infinity is a genuine indeterminate form, and Section 4.8 will finally have the tools to evaluate it properly.
Section
Section 3
Concept
Since an exponential is one-to-one, it has an inverse, and that inverse is the logarithm to the same base. The logarithm of x to base b is the exponent to which b must be raised to give x.
logarithm — For a positive base b other than one, the logarithm to base b of x is the unique exponent y for which b to the power y equals x. The natural logarithm is the one with base e, written ln.
\[ y = \log_b x \iff b^{y} = x \]
Domain and range swap, exactly as Section 1.4 requires. The exponential accepts every real number and outputs only positives, so the logarithm accepts only positives and outputs every real number. That is why the logarithm of a negative number is undefined.
Figure (svg): The exponential and the logarithm reflected in the line y equals x
OpenStax Calculus Volume 1, §1.5 Exponential and Logarithmic Functions §1.5, pp. 87-91 — logarithmic functions
Picture it
The natural exponential and the natural logarithm.
Figure (svg): The exponential and the logarithm reflected in the line y equals x
The point zero comma one on the exponential becomes one comma zero on the logarithm. The exponential's horizontal asymptote becomes the logarithm's vertical asymptote, which is the reflected picture of a function that never reaches zero.
Worked example
Example 1.40. The two forms say the same thing.
\[ \text{Write } \log_3 81 = 4 \text{ in exponential form, and } 2^{5} = 32 \text{ in logarithmic form.} \]
Read the logarithmic form as a question
Why: To what power must 3 be raised to give 81?
\[ \text{the answer is } 4 \]
Write the exponential form
Why: Base to the answer equals the input.
\[ 3 ^{4} = 81 \]
Now go the other way
Why: In the exponential form, the exponent is the answer.
\[ \text{the exponent is } 5 \]
Write the logarithmic form
Why: Log of the result, base 2, equals the exponent.
\[ \log _{2} 32 = 5 \]
Figure (svg): The solution to Worked example converting between forms shown as a ladder of expressions, one row per legal move
\[ 3^4 = 81, \qquad \log_2 32 = 5 \]
Verify: say each aloud as an exponent question
Why: The logarithm base 3 of 81 asks what power of 3 gives 81, and since 3, 9, 27, 81 is four steps the answer is 4. The logarithm base 2 of 32 asks what power of 2 gives 32, and 2, 4, 8, 16, 32 is five steps, so 5. Reading the notation as a question rather than a symbol is what makes these immediate, and it prevents the common confusion of which number is the base.
OpenStax Calculus Volume 1, §1.5 Exponential and Logarithmic Functions §1.5, pp. 88-89
Matching
Each is the other, rearranged.
Match the pairs
Why: The last shows that a logarithm is negative exactly when its input lies between 0 and 1 — the output is still a perfectly ordinary real number, it is only the INPUT that must be positive. Confusing those two restrictions is common and worth guarding against.
Worked example
Checkpoint 1.40. Only positive inputs are allowed.
\[ \text{Find the domain of } f(x) = \ln(3x - 6). \]
Demand a positive argument
Why: The logarithm's domain is the positive numbers.
\[ 3 x - 6 > 0 \]
Solve the inequality
Why: Adding 6 and dividing by 3.
\[ x > 2 \]
Write in interval notation
Why: Strictly greater, so a round bracket.
\[ D = (2, \infty) \]
Note the vertical asymptote
Why: At the excluded endpoint.
\[ \text{asymptote at } x = 2 \]
Figure (svg): The solution to Worked example domain of a logarithmic function shown as a ladder of expressions, one row per legal move
\[ D = (2, \infty) \]
Verify: test the endpoint and a point outside
Why: At x equal to 2 the argument is 0, and the logarithm of 0 is undefined — so the bracket must be round, not square. At x equal to 1 the argument is negative 3, also outside the domain. The strictness matters here in a way it did not for square roots: an even root accepts zero, a logarithm does not, and that single difference is why one gets a square bracket and the other a round one.
OpenStax Calculus Volume 1, §1.5 Exponential and Logarithmic Functions §1.5, pp. 90-91
Trap
\[ \ln(x^2) = 2\ln x \text{ for every real } x \]
Apply the power law without checking the domain
Why: The student moves the exponent out regardless of the sign of x.
\[ \text{at } x = -3: \quad \ln 9 = 2\ln(-3) \quad \text{(right side undefined)} \]
The left side is a perfectly good number, about 2.197, while the right side does not exist at all.
\[ \ln(x^2) = 2\ln|x| \quad \text{for every } x \ne 0 \]
Insert the absolute value the law quietly needs
Why: The power law as usually written assumes a positive argument, and x squared is positive even when x is not.
This is the same absolute-value correction that appeared in Section 1.4 for the root of a square, and it arises for the same reason: an even power destroys the sign, so undoing it must restore the ambiguity. It matters again in Section 5.6, where the antiderivative of one over x is the logarithm of the ABSOLUTE value of x.
Fill the middle
The logarithm from the worked example, which needs a positive argument.
Fill in the blanks
3x - 6 > 0 \;\Longrightarrow\; x > 2
Why: The argument is positive exactly when x exceeds 2, so the domain is the open interval from 2 to infinity. The endpoint is excluded because the logarithm of zero is undefined, and that exclusion is the graph's vertical asymptote.
Sorting
The logarithm accepts only positive inputs; its outputs may be anything.
Sort into buckets
Sort each expression.
The second is the instructive case: the logarithm of 0.2 is about negative 1.61 — a perfectly good negative OUTPUT from a positive INPUT. The restriction is entirely on what goes in, never on what comes out.
Prediction
Commit before reasoning.
Predict first
Why has ln(-4) no real value?
Correct: Because e raised to any real power is positive, so no exponent gives negative 4.
\[ e^{y} > 0 \text{ for every real } y \;\Longrightarrow\; e^{y} = -4 \text{ has no solution} \]
Why: The logarithm asks which exponent produces the input, and the exponential's range is only the positive numbers. Asking for the logarithm of a negative number is asking for a solution to an equation that has none — this is exactly the domain-range swap of Section 1.4, where the inverse's domain must be the original's range. Being a perfect power is irrelevant: the logarithm of 5 exists and is irrational.
Section
Section 4
Concept
Because logarithms are exponents, the rules for combining them are the rules for combining exponents, restated. A product becomes a sum, a quotient becomes a difference, and a power becomes a multiplier.
laws of logarithms — For positive x and y and any real r: the logarithm of a product is the sum of the logarithms, the logarithm of a quotient is their difference, and the logarithm of a power is the exponent times the logarithm.
\[ \log_b(xy) = \log_b x + \log_b y, \quad \log_b(x^{r}) = r\log_b x \]
There is no law for the logarithm of a sum. The logarithm of x plus y is simply not expressible in terms of the logarithms of x and y, and inventing such a rule is the single most common error on this topic.
Figure (svg): The three laws of logarithms, each shown as the exponent rule it comes from
OpenStax Calculus Volume 1, §1.5 Exponential and Logarithmic Functions §1.5, pp. 91-95 — properties of logarithms and change of base
Picture it
Each law beside the exponent rule it comes from.
Figure (svg): The three laws of logarithms, each shown as the exponent rule it comes from
Turning multiplication into addition is what logarithms were invented for, four centuries ago, and it remains their most useful property — including in calculus, where it is the basis of logarithmic differentiation in Section 3.9.
Worked example
Example 1.44. The unknown is upstairs, so bring it down.
\[ \text{Solve } 5^{x} = 30. \]
Take the natural logarithm of both sides
Why: Any base works; the natural one is conventional.
\[ \ln(5 ^{x}) = \ln 30 \]
Apply the power law to bring the exponent down
Why: This is the step the whole method exists for.
\[ x \ln 5 = \ln 30 \]
Solve for the unknown
Why: Divide by the constant.
\[ x = \frac{\ln 30}{\ln 5} \]
Evaluate
Why: About 3.4012 over 1.6094.
\[ x = 2.113 \]
Figure (svg): Solving an exponential equation by taking logarithms of both sides
\[ x = \frac{\ln 30}{\ln 5} \approx 2.113 \]
Verify: bracket the answer between whole powers
Why: Five squared is 25 and 5 cubed is 125, and 30 sits between them much nearer the lower end, so the answer must be a little over 2 — and 2.113 is. This bracketing check also catches the commonest slip, which is inverting the fraction: log 5 over log 30 would be 0.473, far outside the bracket and immediately wrong.
OpenStax Calculus Volume 1, §1.5 Exponential and Logarithmic Functions §1.5, pp. 93-94
Two truths and a lie
All three are about the logarithm laws.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Testing at x equal to y equal to 1 gives ln 2 on the left and 0 on the right. The pattern is seductive because the true law also produces a sum of logarithms — but that sum corresponds to a PRODUCT inside, not a sum. There is simply no law for the logarithm of a sum.
Worked example
Example 1.45. Combine into one logarithm, then undo it.
\[ \text{Solve } \ln(x) + \ln(x - 3) = \ln 10. \]
Combine the left side with the product law
Why: A sum of logarithms is the logarithm of the product.
\[ \ln(x(x - 3)) = \ln 10 \]
Use that the logarithm is one-to-one
Why: Equal logarithms force equal arguments.
\[ x(x - 3) = 10 \]
Solve the resulting quadratic
Why: Expand and factor.
\[ x ^{2} - 3 x - 10 = 0,\text{ so } (x - 5) (x + 2) = 0 \]
Reject any root outside the domain
Why: The original needs x positive AND x minus 3 positive.
\[ x = 5\text{ only} \]
Figure (svg): The solution to Worked example solving a logarithmic equation shown as a ladder of expressions, one row per legal move
\[ x = 5 \]
Verify: substitute both candidate roots
Why: At x equal to 5: ln 5 plus ln 2 is ln 10, which is the right side exactly. At x equal to negative 2: the original has ln of negative 2, which does not exist, so that root is extraneous. Combining logarithms ENLARGES the domain — the product x times x minus 3 is positive for x below 0 as well — so checking every root against the ORIGINAL equation is mandatory, not optional.
OpenStax Calculus Volume 1, §1.5 Exponential and Logarithmic Functions §1.5, pp. 94-95
Error analysis
A student simplifies an expression.
Annotate
On: \( \ln(x + y) = \ln x + \ln y \)
This is the most common error involving logarithms, and it is worth a permanent guard: the sum on the RIGHT of the product law corresponds to a product on the left, never to a sum. When in doubt, test the claim at x equal to y equal to 1, which takes five seconds and settles it.
Fill the middle
Solving the exponential equation from the worked example.
Fill in the blanks
\ln(5^\ln 5) = \ln 30 \;\Longrightarrow\; x\ln 5 = \ln 30 \;\Longrightarrow\; x = \frac______}
Why: The power law converts the unknown exponent into a coefficient, and dividing isolates x. The result is the logarithm of 30 to base 5, written through natural logarithms — which is exactly the change of base formula appearing on its own.
Ranking
Solving a logarithmic equation.
Put in order
Why: Steps a and e bracket the whole method and are both about the same thing: combining logarithms enlarges the domain, so roots can appear that solve the combined equation but not the original. Skipping step e is what leaves an extraneous root in the answer.
Prediction
Commit before reasoning.
Predict first
Why can solving a logarithmic equation produce a root that does not work?
Correct: Because combining logarithms enlarges the domain.
\[ \text{original needs } x > 3; \quad \text{combined needs only } x(x-3) > 0 \]
Why: The original expression needed x positive and x minus 3 positive, so x above 3. The combined form needs only the PRODUCT to be positive, which is also true for x below 0 — so the combined equation has a solution the original never could. The logarithm certainly is one-to-one, which is what justified equating the arguments in the first place. The lesson generalises: any step that widens a domain obliges you to check the answers against the original.
Section
Section 5
Concept
Any logarithm can be computed from any other by dividing, so a calculator with only natural logarithms can produce every base. The same exponential also builds the hyperbolic functions, which are named for their resemblance to the trigonometric ones.
hyperbolic sine and cosine — The hyperbolic cosine is half the sum of e to the x and e to the negative x; the hyperbolic sine is half their difference. They satisfy an identity like the Pythagorean one but with a minus sign, and the hanging-chain curve is a hyperbolic cosine.
\[ \log_b x = \frac{\ln x}{\ln b}, \qquad \cosh x = \frac{e^{x}+e^{-x}}{2} \]
The hyperbolic identity is cosh squared minus sinh squared equals one — a minus where the trigonometric version has a plus. That single sign is the difference between a circle and a hyperbola, and it is where the names come from.
Figure (svg): The change of base formula, with one example evaluated two ways
OpenStax Calculus Volume 1, §1.5 Exponential and Logarithmic Functions §1.5, pp. 92-98 — change of base and hyperbolic functions
Picture it
The hyperbolic cosine and sine.
Figure (svg): The hyperbolic sine and cosine built from the exponential, with the catenary shape
The hyperbolic cosine is even and never drops below one; the hyperbolic sine is odd and passes through the origin. Both are unbounded, which is the clearest difference from their trigonometric namesakes.
Worked example
Example 1.43. Divide, and check the bracket.
\[ \text{Evaluate } \log_2 10 \text{ using natural logarithms.} \]
Write the change of base formula
Why: Log of the input over log of the base, in any common base.
\[ \log _{2} 10 = \frac{\ln 10}{\ln 2} \]
Evaluate the two logarithms
Why: About 2.3026 and 0.6931.
\[ \frac{2.3026}{0.6931} \]
Divide
Why: The quotient.
\[ \text{about } 3.3219 \]
Check by bracketing
Why: Two cubed is 8 and 2 to the fourth is 16.
\[ \text{between } 3\text{ and } 4,\text{ as expected} \]
Figure (svg): The solution to Worked example change of base shown as a ladder of expressions, one row per legal move
\[ \log_2 10 = \frac{\ln 10}{\ln 2} \approx 3.3219 \]
Verify: raise the base to the answer
Why: Two to the 3.3219 is about 10.00, confirming the value directly. The bracketing check is the faster one though: since 10 lies between 8 and 16, the answer must lie between 3 and 4, and this catches the inverted fraction — ln 2 over ln 10 gives 0.301, nowhere near the bracket.
OpenStax Calculus Volume 1, §1.5 Exponential and Logarithmic Functions §1.5, pp. 92-93
Fill the middle
Evaluating a base-two logarithm with natural logarithms.
Fill in the blanks
\log_2 10 = \frac\ln 2___} \approx 3.3219
Why: The base goes in the denominator and the argument on top. The bracketing check confirms it: the answer must lie between 3 and 4 because 10 lies between 2 cubed and 2 to the fourth.
Worked example
Example 1.47. Verify it from the definitions.
\[ \text{Show that } \cosh^{2} x - \sinh^{2} x = 1. \]
Write both in terms of the exponential
Why: Half sum and half difference.
\[ (\frac{e ^{x} + e ^{-x}}{2}) ^{2} - (\frac{e ^{x} - e ^{-x}}{2}) ^{2} \]
Expand each square
Why: The cross terms have opposite signs.
\[ \frac{e ^{2} x + 2 + e ^{-2} x}{4} - \frac{e ^{2} x - 2 + e ^{-2} x}{4} \]
Subtract
Why: The squared terms cancel entirely.
\[ \frac{2 + 2}{4} \]
Simplify
Why: Four quarters.
\[ = 1 \]
Figure (svg): The solution to Worked example the hyperbolic identity shown as a ladder of expressions, one row per legal move
\[ \cosh^{2}x - \sinh^{2}x = 1 \]
Verify: test at a convenient input
Why: At x equal to 0: cosh 0 is 1 and sinh 0 is 0, so the left side is 1 minus 0, which is 1. At x equal to 1: cosh 1 is about 1.5431 and sinh 1 about 1.1752, and 2.3812 minus 1.3811 is 1.0001, which is 1 up to rounding. Notice the MINUS sign, where the trigonometric identity has a plus — a point on the trigonometric circle satisfies x squared plus y squared equals 1, while the hyperbolic pair satisfies x squared minus y squared equals 1, which is a hyperbola. That is exactly where the name comes from.
OpenStax Calculus Volume 1, §1.5 Exponential and Logarithmic Functions §1.5, pp. 96-97
Trap
\[ \log_2 10 = \frac{\ln 2}{\ln 10} \approx 0.301 \quad \text{(wrong)} \]
Put the base on top
Why: The student writes the two logarithms in the order the symbols appear.
But 2 raised to the power 0.301 is about 1.23, not 10. The answer is not even close.
\[ \log_2 10 = \frac{\ln 10}{\ln 2} \approx 3.3219 \]
The ARGUMENT goes on top, the base underneath
Why: The formula converts the question 'what power of 2 gives 10' and the thing being reached for is 10.
The bracketing check settles it without remembering the order at all. Since 2 cubed is 8 and 2 to the fourth is 16, the answer must be between 3 and 4. Only one of the two fractions lands there, so computing both and keeping the plausible one is a legitimate strategy.
Matching
The identities rhyme, but not exactly.
Match the pairs
Why: The parities match their trigonometric namesakes exactly, which is why the names were chosen. The identity differs in one sign, and that sign is the whole distinction between the two families: plus gives a circle, minus gives a hyperbola.
Sorting
Compare each with its trigonometric namesake.
Sort into buckets
Sort each function.
This is the sharpest difference between the two families. The trigonometric functions are periodic and bounded; the hyperbolic ones are neither. The shared names describe an algebraic resemblance, not a behavioural one.
Prediction
Commit before reasoning.
Predict first
A chain hangs freely between two posts. What curve does it form?
Correct: A hyperbolic cosine — the catenary.
\[ y = a\cosh\!\left(\tfrac{x}{a}\right) \quad \text{the catenary} \]
Why: The shape looks very like a parabola and was believed to be one for a long time, but solving the physics gives a hyperbolic cosine. The distinction is real though subtle: a catenary is slightly flatter at the bottom and rises more steeply at the sides. Galileo guessed a parabola; it took the calculus of the seventeenth century to settle it, and the answer is one of the first genuine applications of the function family defined here. Section 6.4 computes the length of such a curve.
Comparison
Fill the blanks. Everything in the table follows from their being inverses.
Comparison matrix
| Property | b to the x | log base b of x |
|---|---|---|
| Domain | all real numbers | the positive numbers |
| Range | the positive numbers | all real numbers |
| Passes through | (0, 1) | (1, 0) |
| Asymptote | horizontal, the x-axis | vertical, the y-axis |
Every row is the previous column reflected in the line y equals x. That is not a coincidence to memorise but the definition of an inverse from Section 1.4, applied once.
Pattern
Given an equation with the unknown in an exponent or inside a logarithm.
Step five is not a formality. Combining logarithms genuinely enlarges the domain, so extraneous roots are produced by a correct method rather than by a mistake, and only the check removes them.
Stewart, Calculus: Early Transcendentals 8e, §1.4 Exponential Functions §1.4, pp. 45-54
Check
Modelling. The base is what remains.
Check your understanding
A quantity of 80 mg falls by 30 percent each hour. What is the model?
Answer: A
Why: Losing 30 percent leaves 70 percent, so the multiplier is 0.7.
Check
Solving. Bring the exponent down.
Check your understanding
Solve 5^x = 30.
Answer: A
Why: Taking logarithms gives x times ln 5 equals ln 30, so x is their quotient.
Check
Extraneous roots. Check against the original.
Check your understanding
Solve ln x + ln(x - 3) = ln 10.
Answer: A
Why: The quadratic gives 5 and negative 2, but negative 2 makes both logarithms undefined.
Real world
Carbon-14 decays with a half-life of about 5730 years. A sample from an archaeological site retains 23 percent of the carbon-14 a living organism would have.
Discussion prompt
Build the decay model, find the continuous decay rate, and estimate the sample's age. Then say why the answer is quoted with only two significant figures.
Hint: The half-life gives one equation; solving it for the rate is the whole first half of the problem.
Answer:
\[ N(t) = N_0 e^{-kt}, \qquad \tfrac{1}{2} = e^{-5730k} \]
\[ -5730k = \ln\tfrac{1}{2} = -\ln 2 \;\Longrightarrow\; k = \frac{\ln 2}{5730} \approx 1.2097 \times 10^{-4} \]
Now use the measured fraction. Setting the ratio to 0.23 and taking logarithms:
\[ 0.23 = e^{-kt} \;\Longrightarrow\; t = -\frac{\ln 0.23}{k} = \frac{1.4697}{1.2097 \times 10^{-4}} \approx 12{,}150 \text{ years} \]
So the sample is roughly 12,000 years old — a little over two half-lives, which is the sanity check: two half-lives would leave 25 percent, and 23 percent is slightly less, so slightly more than two half-lives. That bracket confirms the arithmetic without repeating it.
Two significant figures is honest because the input had two. A measurement of 23 percent could be anywhere from 22.5 to 23.5, which moves the answer by about 180 years either way — so quoting 12,150 would claim a precision the data does not support.
Commit first
Answer, then rate your confidence honestly.
Predict first
Which of these is a genuine law of logarithms?
Correct: The power law: the logarithm of a power is the exponent times the logarithm.
\[ \log_b(x^{r}) = r\log_b x \quad \text{because } (b^{m})^{r} = b^{mr} \]
Why: The power law is genuine and is the one that makes exponential equations solvable at all. The first is the classic false rule — testing at x equal to y equal to 1 gives ln 2 against 0. The third confuses dividing two logarithms with the logarithm of a quotient, which is a difference, not a division. The fourth turns a product into a product of logarithms, where the true law gives a sum. Each false option is a real pattern applied to the wrong operation, which is exactly why testing at a small input is worth doing whenever a law feels uncertain.
Explain it
They can compute logarithms on a calculator but think of them as an unrelated button.
Discussion prompt
In four sentences or fewer, explain what a logarithm is and why it can only accept positive inputs.
Hint: State it as a question about exponents.
Answer:
A logarithm answers the question: what exponent do I put on the base to get this number? The logarithm base 2 of 8 is 3 because 2 cubed is 8, and that is all the notation is saying.
Now ask the same question about a negative input: what power of 2 gives negative 8? There is none — every real power of a positive base comes out positive. So the logarithm of a negative number is not merely hard to compute, it does not exist, and that is why the domain is the positive numbers only.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For modelling, always write the base as one plus or minus the rate, so the direction is explicit. For the laws, test any doubtful one at x equal to y equal to 1 — it takes five seconds and settles it. For exponential equations, take logarithms of both sides immediately and let the power law do the work, then bracket the answer between whole powers. For extraneous roots, write the domain condition down before you start and check every root against it. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
On one set of axes with the diagonal dashed, draw the natural exponential and the natural logarithm as mirror images, marking the point zero comma one on one and one comma zero on the other, and label the horizontal asymptote of one and the vertical asymptote of the other. Beside it, write the compound interest table from one compounding per year to continuous, and circle the number it converges to. In the middle of the page write the three laws of logarithms, and beside each the exponent law it comes from; then write the false law about the logarithm of a sum, cross it out, and put the counterexample at x equal to y equal to 1 beside it. Below that, solve 5 to the x equals 30 in full, and solve ln x plus ln of x minus 3 equals ln 10 in full, showing the extraneous root and one line saying why it must be discarded. In a margin, write the definitions of the hyperbolic sine and cosine and the identity relating them, noting the sign that differs from the trigonometric case.
If your exponential and logarithm graphs are not mirror images across the dashed diagonal, check a single pair: whatever point (a, b) sits on one, (b, a) must sit on the other. The pair (0, 1) and (1, 0) is the fastest one to test.
Recap
Five things, and the second is why the rest of this course writes e rather than any other base.
| If you see | Then |
|---|---|
| A percentage rate | The base is one plus or minus it |
| Continuous compounding | Use e to the rate times time |
| The unknown in an exponent | Take logarithms and use the power law |
| A sum of logarithms | Combine into the logarithm of a product |
| The logarithm of a sum | It does not simplify at all |
| A solved logarithmic equation | Check every root against the original |
| An unfamiliar base | Divide two natural logarithms |
That completes Chapter 1. Chapter 2 stops cataloguing functions and asks the question calculus is built on: what does a function do NEAR a point, as opposed to AT one — and Section 2.1 shows why two ancient geometric problems both need that same idea.
OpenStax Calculus Volume 1, §1.5 Exponential and Logarithmic Functions §1.5, pp. 80-98 — everything on these slides traces back here
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