Writing an exponential function through two points, the semi-log transformation that straightens exponential data, finding an exponential model from a scatter plot and by regression, writing a power function through two points, and the log-log transformation that straightens power data.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 7 — Exponential and Logarithmic Functions
Write and Apply Exponential and Power Functions
Objectives
Five outcomes. Two families, and one transformation each that turns a curve into a line.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.7 Write and Apply Exponential and Power Functions §7.7, pp. 528-535 — the lesson these objectives are drawn from
Warm-up
Chapter 2 fitted a line to two points. Chapter 4 fitted a parabola. This lesson fits the last two families.
Discussion prompt
A line has two constants, slope and intercept, so two points determine it. How many constants does the exponential family y equals a times b to the x have, and how many points should it need?
Hint: Count the letters that are not x or y.
Answer:
Two constants, a and b, so two points determine an exponential curve — exactly as two points determine a line.
\[ y = ab^x \quad \text{and} \quad y = ax^b \quad \text{each have two constants} \]
The power family has two as well. So both can be pinned down from a pair of points, and the only question is what algebra gets you there.
Concept
Exponential data becomes straight when the natural logarithm of y is plotted against x. Power data becomes straight when the natural logarithm of y is plotted against the natural logarithm of x. Either way, a line can be fitted and then undone.
power function — A function of the form y equals a times x to the b, where the variable is in the base and the exponent is a constant. In an exponential function it is the other way round.
\[ y = ab^x \;\Longleftrightarrow\; \ln y = (\ln b)x + \ln a \]
This is why logarithmic graph paper existed long before regression keys did: a straight line can be drawn and measured by hand, and a curve cannot.
Figure (svg): Two columns comparing the semi-log and log-log transformations
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.7 Write and Apply Exponential and Power Functions §7.7, pp. 529-532
Section
Section 1
Concept
Substitute both points into y equals a times b to the x, solve the first equation for a, substitute into the second, and the a cancels. What is left determines b, and back-substituting gives a.
\[ 12 = ab^1, \; 108 = ab^3 \;\Longrightarrow\; b = 3, \; a = 4 \]
The base b must be positive, so only the positive square root is taken. Two constants need two equations, which is exactly what two points supply.
Figure (svg): An exponential function found from two points on its graph
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.7 Write and Apply Exponential and Power Functions §7.7, pp. 529-529 — Write an exponential function
Picture it
Example 1: through the points where x is 1 and 3.
Figure (svg): An exponential function found from two points on its graph
Dividing the second equation by the first is what removes a, leaving b squared equal to 9. The rest is arithmetic.
Worked example
Example 1.
\[ \text{Write } y = ab^x \text{ whose graph passes through } (1, 12) \text{ and } (3, 108). \]
Substitute both points
Why: Each point gives one equation in a and b.
\[ 12 =\text{ ab and } 108 = a b ^{3} \]
Solve the first for a
Why: Dividing both sides by b isolates a.
\[ a = \frac{12}{b} \]
Substitute into the second
Why: The b in the denominator cancels one factor of b cubed.
\[ 108 = 12 b ^{2} \]
Solve for b, then for a
Why: Nine is b squared, so b is 3, and a is 12 over 3.
\[ b = 3, a = 4 \]
Figure (svg): An exponential function found from two points on its graph
\[ y = 4 \cdot 3^x \]
Verify: substitute both points back
Why: At x equal to 1, 4 times 3 is 12. At x equal to 3, 4 times 27 is 108. Both given points are on the curve. Only the positive root of 9 was used, because an exponential base must be positive — the negative root would not define a function on the reals.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.7 Write and Apply Exponential and Power Functions §7.7, pp. 529-529
Fill the middle
Example 1, second step.
Fill in the blanks
108 = \left(\tfrac12___\right)b^3 = ___b^2
Why: The b in the denominator cancels one factor from b cubed, leaving 12 times b squared. Eliminating a in one step is what makes two points enough to determine the whole curve.
Worked example
Guided Practice 1 to 3.
\[ \text{Write } y = ab^x \text{ through } (1,6),(3,24); \; (2,8),(3,32); \; (3,8),(6,64). \]
First: divide the equations
Why: Twenty-four over 6 is b squared, which is 4.
\[ b = 2, a = 3 \]
Second: the exponents differ by 1
Why: Thirty-two over 8 is b directly.
\[ b = 4, a = 0.5 \]
Third: the exponents differ by 3
Why: Sixty-four over 8 is b cubed, which is 8.
\[ b = 2, a = 1 \]
Note the shortcut
Why: Dividing the second equation by the first always eliminates a in one line.
Figure (svg): The solution to Worked example three more exponential functions shown as a ladder of expressions, one row per algebraic move
\[ y = 3 \cdot 2^x, \quad y = 0.5 \cdot 4^x, \quad y = 2^x \]
Verify: check the second at both points
Why: At x equal to 2, 0.5 times 16 is 8. At x equal to 3, 0.5 times 64 is 32. Both check. Notice how the ratio of the two y values equalled b raised to the difference of the x values — a shortcut worth using, since it skips the substitution entirely.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.7 Write and Apply Exponential and Power Functions §7.7, pp. 531-531
Trap
\[ 9 = b^2 \;\Longrightarrow\; b = 3 \text{ or } b = -3 \]
Keep both roots
Why: Solving b squared equals 9 usually gives two answers.
\[ y = -4(-3)^x \quad \text{(wrong)} \]
A negative base gives no real value at most non-integer exponents. At x equal to one half it would ask for the square root of a negative number.
\[ 9 = b^2 \;\Longrightarrow\; b = 3 \text{, taking the positive root} \]
Discard the negative root on domain grounds
Why: An exponential function requires a positive base, as Lesson 7.1 established.
\[ y = 4 \cdot 3^x \]
This is the same restriction that made the logarithm's base positive in Lesson 7.4, and for the same reason: a negative base has no consistent real powers.
Matching
Divide the y values and read the exponent difference.
Match the pairs
Why: In every row the ratio of the y values equals b raised to the difference of the x values: 9 equals 3 squared, 4 equals 2 squared, 4 equals 4 to the first, and 8 equals 2 cubed. Reading that ratio first gives b before any substitution.
Sorting
Count the constants in each family.
Sort into buckets
Sort each family by how many points determine it.
The count of constants is the count of points, every time. It is the same principle behind Lesson 4.10's three-point parabola and Lesson 2.4's two-point line.
Prediction
Commit before reasoning.
Predict first
Guided Practice 8 asks for a power function through (3, 5) and (3, 7). What happens?
Correct: The method fails, because no function has two outputs at one input.
\[ \tfrac{7}{5} = \left(\tfrac{3}{3}\right)^b = 1^b = 1, \text{ impossible} \]
Why: Dividing the two equations gives 7 over 5 equal to 1 raised to b, which says 1.4 equals 1 — impossible. Algebra is reporting what the vertical line test already said in Lesson 2.1: a relation with two different outputs at the same input is not a function at all. The same collapse would happen for an exponential fit through two points sharing an x value.
Section
Section 2
Concept
A set of more than two points fits an exponential pattern exactly when the transformed points, with the natural logarithm of y in place of y, fall on a line. Taking a logarithm of both sides of the model shows why.
\[ y = ab^x \;\Longrightarrow\; \ln y = (\ln b)x + \ln a \]
The transformed equation is a line in x with slope equal to the natural logarithm of b and intercept the natural logarithm of a. Straightness of the transformed plot is therefore the test for an exponential fit.
Figure (svg): An exponential curve beside the straight line its natural logarithms produce
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.7 Write and Apply Exponential and Power Functions §7.7, pp. 529-529 — Transforming exponential data
Picture it
The same five points, plotted twice.
Figure (svg): An exponential curve beside the straight line its natural logarithms produce
On the left they curve upward; on the right they lie exactly on a line of slope the natural logarithm of 2, which is about 0.69.
Worked example
Example 2, Steps 1 and 2.
\[ \text{Scooter sales in years 1 to 7 were } 12, 16, 25, 36, 50, 67, 96. \text{ Test for an exponential pattern.} \]
Take the natural logarithm of each sales figure
Why: Twelve becomes 2.48, 16 becomes 2.77, and so on.
\[ 2.48, 2.77, 3.22, 3.58, 3.91, 4.20, 4.56 \]
Plot the transformed pairs
Why: The horizontal coordinate stays the year.
\[ (1, 2.48)\text{ up to } (7, 4.56) \]
Look at the shape
Why: The points lie close to a straight line.
Conclude
Why: Straight after transforming means exponential before transforming.
Figure (svg): A scatter plot of scooter sales after taking the natural logarithm of the sales figures
\[ \text{(x, ln y) is nearly linear, so } y = ab^x \text{ fits} \]
Verify: check the successive differences
Why: The transformed values rise by 0.29, 0.45, 0.36, 0.33, 0.29 and 0.36 — all close to one another, which is what a constant slope means. Equivalently, the original sales figures multiply by roughly 1.4 each year rather than adding a fixed amount, which is the defining behaviour of exponential growth.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.7 Write and Apply Exponential and Power Functions §7.7, pp. 530-530
Fill the middle
Example 2, Step 3.
Fill in the blanks
m = \frac0.35___ = ___
Why: The slope is about 0.35, and it equals the natural logarithm of the growth factor b. Exponentiating it gives b directly: e to the 0.35 is about 1.42.
Worked example
Example 2, Step 3.
\[ \text{Fit a line through } (1, 2.48) \text{ and } (7, 4.56) \text{ and convert it back to } y = ab^x. \]
Find the slope
Why: Two point zero eight over 6.
\[ \text{about } 0.35 \]
Write the line in point-slope form
Why: The vertical axis is the natural logarithm of y, not y.
\[ \ln y - 2.48 = 0.35(x - 1) \]
Simplify and exponentiate
Why: The line becomes ln y equals 0.35x plus 2.13.
\[ y = e ^{0.35 x + 2.13} \]
Split the exponent
Why: The product of powers rule separates the constant from the variable part.
\[ y = e ^{2.13}(e ^{0.35}) ^{x} \]
Figure (svg): A scatter plot of scooter sales after taking the natural logarithm of the sales figures
\[ y = 8.41(1.42)^x \]
Verify: test the model at both fitted points
Why: At x equal to 1, 8.41 times 1.42 is 11.9, close to the actual 12. At x equal to 7, 1.42 to the seventh is 11.6, and 8.41 times that is 97.7, close to the actual 96. Splitting the exponent is the step that converts a line back into an exponential, and it is nothing but the product of powers rule from Lesson 5.1.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.7 Write and Apply Exponential and Power Functions §7.7, pp. 530-530
Error analysis
A student tests some data for an exponential pattern.
Annotate
On: \( \text{plotted } (\ln x, y) \text{ instead of } (x, \ln y) \)
Decide which variable is trapped by the exponent first. In an exponential model that is y; in a power model both variables need transforming.
Comparison
Fill the blanks. The transformed line carries both constants.
Comparison matrix
| On the transformed line | Means | Here |
|---|---|---|
| slope | ln b | 0.35, so b is about 1.42 |
| intercept | ln a | 2.13, so a is about 8.41 |
| straightness | the data really is exponential | close, so a good fit |
| positive slope | b greater than 1, so growth | sales are rising |
Every feature of the transformed line translates back into a feature of the original model, which is what makes the detour through logarithms worth taking.
Prediction
Commit before reasoning.
Predict first
You plot the points with the natural logarithm of y and they curve. What follows?
Correct: The data is not exponential; some other family may still fit.
\[ (x, \ln y) \text{ linear} \;\Longleftrightarrow\; y = ab^x \]
Why: The test is an if-and-only-if: straight after the transformation means exponential, and curved means not exponential. It says nothing about which family DOES fit — a power model, a quadratic or a logarithmic model might all still work, and the log-log plot of the next idea is the natural thing to try next. A negative result narrows the search rather than ending it.
Sorting
Look at how the y values change.
Sort into buckets
Sort each sequence of y values at x equal to 1, 2, 3, 4.
The third one multiplies by about 1.4 each step rather than exactly, which is what real data looks like — close enough for the transformed plot to be nearly straight.
Section
Section 3
Concept
Fitting a line by eye through two chosen points uses only those two. A calculator's exponential regression uses all the data to find the best-fitting model, and it is the tool to reach for once the transformed plot has confirmed the family.
\[ y = 8.46(1.42)^x \]
The hand-fitted model and the regression model here agree closely, 8.41 against 8.46, which is a sign that the two chosen points were representative of the whole set.
Figure (svg): A scatter plot of scooter sales after taking the natural logarithm of the sales figures
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.7 Write and Apply Exponential and Power Functions §7.7, pp. 530-530 — Use exponential regression
Picture it
The same scooter data, fitted by hand and by regression.
Figure (svg): A scatter plot of scooter sales after taking the natural logarithm of the sales figures
The hand fit gave 8.41 and 1.42; the regression gave 8.46 and 1.4188. Close agreement means the two chosen points sat well on the trend.
Worked example
Example 3.
\[ \text{Using } y = 8.46(1.42)^x, \text{ predict scooter sales in the eighth year.} \]
Identify the input
Why: The eighth year means x equal to 8.
\[ x = 8 \]
Raise the base to that power
Why: One point four two squared is 2.016, to the fourth is 4.07, to the eighth is 16.5.
\[ 1.42 ^{8} = 16.53 \]
Multiply by the initial value
Why: Eight point four six times 16.53.
\[ \text{about } 140 \]
State the answer in context
Why: Sales are a whole number of scooters.
\[ \text{about } 140\text{ scooters} \]
Figure (svg): The solution to Worked example predict with a regression model shown as a ladder of expressions, one row per algebraic move
\[ y = 8.46(1.42)^8 \approx 140 \]
Verify: check the growth against year 7
Why: Year 7 actually saw 96 scooters, and multiplying by the growth factor 1.42 gives 136 — close to the model's 140. The prediction is one step beyond the data, which is the safest kind of extrapolation. Predicting year 20 would give over 26,000 scooters, which no shop sells, so the model's reach has limits.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.7 Write and Apply Exponential and Power Functions §7.7, pp. 530-530
Fill the middle
Example 3.
Fill in the blanks
y = 8.46(1.42)^8} \approx 140
Why: The eighth year is x equal to 8, so the exponent is 8. Because the years were numbered from 1 rather than 0, no shifting is needed — but always check which year the table calls x equal to 1.
Worked example
Guided Practice 4.
\[ \text{Sales of } 15, 23, 40, 52, 80, 105, 140 \text{ in years 1 to 7. How does the model change?} \]
Transform the data
Why: The natural logarithms run 2.71, 3.14, 3.69, 3.95, 4.38, 4.65, 4.94.
Find the slope
Why: Two point two three over 6.
\[ \text{about } 0.372 \]
Write the line and exponentiate
Why: The intercept is about 2.34.
\[ y = e ^{2.34}(e ^{0.372}) ^{x} \]
Read off the constants
Why: The initial value rises and so does the growth factor.
\[ y\text{ about } 10.3(1.45) ^{x} \]
Figure (svg): The solution to Worked example a different sales table shown as a ladder of expressions, one row per algebraic move
\[ y \approx 10.3(1.45)^x \]
Verify: compare the two models
Why: The first model was 8.41 times 1.42 to the x, and this one is 10.3 times 1.45 to the x. Both the starting level and the growth rate went up, which matches a table whose every entry is larger and whose ratios are slightly bigger. A model that changed in only one constant would have been suspicious.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.7 Write and Apply Exponential and Power Functions §7.7, pp. 531-531
Trap
\[ y = 8.46(1.42)^x \text{ at } x = 25 \]
Substitute and report the answer
Why: The model is treated as valid for any year.
\[ y \approx 8.46(1.42)^{25} \approx 22{,}000 \quad \text{(meaningless)} \]
No single shop sells 22,000 scooters a year. The model was fitted to seven years and describes those years, not the next twenty-five.
\[ y = 8.46(1.42)^8 \approx 140 \]
Predict one step beyond the data, and say so
Why: A model earns trust only over the range it was fitted to and a little past it.
\[ \text{years 1 to 7 fitted; year 8 predicted} \]
Every exponential model of a real quantity eventually meets a limit — a market size, a food supply, a physical constraint. Lesson 7.2's decay curves flatten toward an asymptote; growth curves in the real world flatten too, just not within this model.
Ranking
From raw data to a prediction.
Put in order
Why: Step two is the one that is easiest to skip and most expensive to skip: fitting an exponential to data that is not exponential produces a model that looks respectable and predicts badly. The straightness check is what licenses everything after it.
Two truths and a lie
All three are about fitting exponential models.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. A model describes the range it was fitted to; extrapolating an exponential far past that range grows absurdly fast. The scooter model fitted seven years beautifully and predicts 22,000 sales in year 25. Fit quality and extrapolation reach are different questions, and only the first is measured by the data.
Matching
Read the two constants.
Match the pairs
Why: The last row is the link between the two pictures: the transformed line's slope is not b itself but its natural logarithm, which is why exponentiating is needed to recover the model. A negative slope there would mean b below 1 and therefore decay.
Section
Section 4
Concept
A power function has the form y equals a times x to the b, with the variable in the base. Two points give two equations, and eliminating a leaves an exponential equation in b that a logarithm solves.
\[ 2 = a\cdot 3^b, \; 9 = a\cdot 6^b \;\Longrightarrow\; b \approx 2.17, \; a \approx 0.184 \]
The contrast with the previous case is worth noticing. There the unknown was the base and a root finished the job; here the unknown is the exponent and a logarithm does.
Figure (svg): A power function found from two points on its graph
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.7 Write and Apply Exponential and Power Functions §7.7, pp. 531-531 — Write a power function
Picture it
Example 4: through the points where x is 3 and 6.
Figure (svg): A power function found from two points on its graph
Dividing gives 4.5 equal to 2 to the b, since 6 over 3 is 2. From there Lesson 7.6's method finishes it.
Worked example
Example 4.
\[ \text{Write } y = ax^b \text{ whose graph passes through } (3, 2) \text{ and } (6, 9). \]
Substitute both points
Why: Each gives one equation in a and b.
\[ 2 = a \cdot 3 ^{b}\text{ and } 9 = a \cdot 6 ^{b} \]
Solve the first for a and substitute
Why: Six over 3 is 2, so the bases collapse.
\[ 9 = 2 \cdot 2 ^{b} \]
Take a logarithm
Why: Four point five equals 2 to the b, so b is the logarithm base 2 of 4.5.
\[ b = \log 4.5 / \log 2 = 2.17 \]
Back-substitute for a
Why: Two divided by 3 to the 2.17, which is about 10.85.
\[ a = 0.184 \]
Figure (svg): A power function found from two points on its graph
\[ y = 0.184x^{2.17} \]
Verify: substitute both points back
Why: At x equal to 3, 3 to the 2.17 is 10.85, and 0.184 times that is 2.00. At x equal to 6, 6 to the 2.17 is 48.9, and 0.184 times that is 9.00. Both points check to two decimal places, which is all the rounding allows.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.7 Write and Apply Exponential and Power Functions §7.7, pp. 531-531
Comparison
Fill the blanks. The difference is where x sits.
Comparison matrix
| Question | y = ab^x | y = ax^b |
|---|---|---|
| Where is x? | in the exponent | in the base |
| Unknown after eliminating a | the base b, found with a root | the exponent b, found with a logarithm |
| Transformation that straightens | (x, ln y) | (ln x, ln y) |
| Long-run behaviour | outgrows every power | grows, but polynomially |
The last row is why the distinction matters beyond bookkeeping: an exponential eventually overtakes any power function, however large that power is.
Worked example
Guided Practice 5 to 7.
\[ \text{Write } y = ax^b \text{ through } (2,1),(7,6); \; (3,4),(6,15); \; (5,8),(10,34). \]
First: divide the equations
Why: Six over 1 equals 3.5 to the b, so b is about 1.43.
\[ a = 1 / 2 ^{1.43} = 0.371 \]
Second: the x values double
Why: Fifteen over 4 is 3.75 equal to 2 to the b.
\[ b = 1.907 \]
Second: back-substitute
Why: Four divided by 3 to the 1.907, which is about 8.12.
\[ a = 0.492 \]
Third: the x values double again
Why: Thirty-four over 8 is 4.25 equal to 2 to the b, so b is 2.087.
\[ a = 0.278 \]
Figure (svg): The solution to Worked example three more power functions shown as a ladder of expressions, one row per algebraic move
\[ 0.371x^{1.43}, \; 0.492x^{1.907}, \; 0.278x^{2.087} \]
Verify: check the third at its second point
Why: Ten to the 2.087 is about 122.3, and 0.278 times that is 34.0 — the given value. Notice that in the last two the x values doubled, which made the ratio equal to 2 to the b and the logarithm easy. Choosing points whose x values are in a simple ratio always simplifies this step.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.7 Write and Apply Exponential and Power Functions §7.7, pp. 531-531
Error analysis
A student is asked for a power function and writes an exponential one instead.
Annotate
On: \( \text{through } (3,2), (6,9): \; y = ab^x \text{ instead of } y = ax^b \)
Read where x sits before choosing a method. If x is in the exponent it is exponential; if x is in the base with a constant exponent it is a power function.
Fill the middle
Example 4, second step.
Fill in the blanks
9 = \left(\tfrac2___\right)6^b = 2\left(\tfrac______\right)^b = 2 \cdot ___^b
Why: Six over 3 is 2, so the two powers collapse into a single power of 2. Choosing points whose x values are in a simple ratio is what makes this step painless.
Sorting
Find x and see whether it is up or down.
Sort into buckets
Sort each function.
A decimal exponent is a strong hint of a power model, and a decimal base a strong hint of an exponential one — but reading where x sits is the only reliable test.
Prediction
Commit before reasoning.
Predict first
Compare y equals x to the 2.5 with y equals 1.42 to the x for very large x.
Correct: The exponential overtakes it and then leaves it far behind.
\[ 1.42^x \text{ eventually exceeds } x^k \text{ for every fixed } k \]
Why: At x equal to 10 the power function gives 316 and the exponential only 33, so the power leads early. But by x equal to 60 the exponential is over 2 million and the power is under 28,000. Any exponential with a base above 1 eventually outgrows any power function, no matter how large the exponent or how small the coefficients — the crossing point just moves. That is why the distinction between the two families is more than notation.
Section
Section 5
Concept
A set of more than two points fits a power pattern exactly when the transformed points, with the natural logarithm of both coordinates, fall on a line. The slope of that line is the exponent itself.
\[ y = ax^b \;\Longrightarrow\; \ln y = b\ln x + \ln a \]
Only the exponential case leaves x alone. Here the variable is in the base, so it needs the logarithm too — and the payoff is that the slope reads off the exponent directly, with no exponentiating.
Figure (svg): A power curve beside the straight line its double logarithms produce
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.7 Write and Apply Exponential and Power Functions §7.7, pp. 532-532 — Transforming power data
Picture it
The square root function, plotted twice.
Figure (svg): A power curve beside the straight line its double logarithms produce
The right plot is a line of slope one half — exactly the exponent of the original function. The exponent is readable straight off the graph.
Worked example
Example 5, Steps 1 and 2.
\[ \text{For five birds, } (\ln x, \ln y) = (0.642,-1.470), (1.072,0.039), (1.227,0.525), (1.677,1.911), (2.128,2.774). \]
Take the natural logarithm of both variables
Why: Wingspan in feet and weight in pounds are both transformed.
Plot the transformed pairs
Why: The horizontal axis is now the logarithm of wingspan.
Look at the shape
Why: The five points lie close to a straight line.
Conclude
Why: Straight on a log-log plot means a power pattern in the original data.
Figure (svg): Wingspans and weights of five birds, plotted after taking logarithms of both variables
\[ (\ln x, \ln y) \text{ nearly linear, so } y = ax^b \text{ fits} \]
Verify: check the consistency of the slopes
Why: From the first point to the third the rise is 1.995 over a run of 0.585, a slope of 3.41; from the third to the fifth it is 2.249 over 0.901, a slope of 2.50. Real biological data is not perfect, but the points still cluster near one line closely enough for the model to be useful — which is the standard being applied here.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.7 Write and Apply Exponential and Power Functions §7.7, pp. 532-532
Fill the middle
Example 5, Step 3.
Fill in the blanks
m = \frac2.5___ \approx ___ = b
Why: The slope is about 2.5, and on a log-log plot the slope IS the exponent — no exponentiating needed for that constant. Only the intercept has to be exponentiated, to recover the coefficient a.
Worked example
Example 5, Step 3.
\[ \text{Fit a line through } (1.227, 0.525) \text{ and } (2.128, 2.774) \text{ and convert it to } y = ax^b. \]
Find the slope
Why: Two point two four nine over 0.901.
\[ \text{about } 2.5 \]
Write the line in point-slope form
Why: Both axes are logarithms, so x is replaced by ln x.
\[ \ln y - 2.774 = 2.5(\ln x - 2.128) \]
Simplify and use the power property
Why: Two point five times the logarithm of x is the logarithm of x to the 2.5.
\[ \ln y = \ln(x ^{2.5}) - 2.546 \]
Exponentiate and split
Why: E to the negative 2.546 is about 0.0784.
\[ y = 0.0784 x ^{2.5} \]
Figure (svg): Wingspans and weights of five birds, plotted after taking logarithms of both variables
\[ y = 0.0784x^{2.5} \]
Verify: test the model on the largest bird
Why: The largest wingspan has natural logarithm 2.128, so the wingspan is about 8.4 feet. Then 8.4 to the 2.5 is about 204, and 0.0784 times that is 16.0 pounds — matching the natural logarithm 2.774, since e to that is 16.0. The model reproduces the point it was fitted through, as it must.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.7 Write and Apply Exponential and Power Functions §7.7, pp. 532-532
Trap
\[ \ln y = 2.5\ln x - 2.546 \]
Report the model as y equals 2.5x minus 2.546
Why: The line's coefficients are copied straight across.
\[ y = 2.5x - 2.546 \quad \text{(wrong)} \]
That is a line, not a power curve. The equation describes the LOGARITHMS, not the variables themselves.
\[ \ln y = \ln(x^{2.5}) - 2.546 \;\Longrightarrow\; y = e^{-2.546}x^{2.5} \]
Exponentiate to undo both logarithms
Why: The slope becomes the exponent and the intercept becomes the coefficient, after exponentiating.
\[ y = 0.0784x^{2.5} \]
The same warning applies in the semi-log case: there the slope became the logarithm of the growth factor, and the intercept the logarithm of the initial value. The transformed line always has to be undone.
Matching
Which variable carries the logarithm?
Match the pairs
Why: The last two rows are the practical difference between the pictures: a log-log slope is the answer, while a semi-log slope needs one more step. Both intercepts, however, are logarithms of the coefficient and both need exponentiating.
Prediction
Commit before reasoning.
Predict first
The model gives weight as roughly the 2.5 power of wingspan. What would an exponent near 3 have meant?
Correct: That birds scale like solid objects, with weight proportional to length cubed.
\[ \text{geometric scaling: } y \propto x^3; \quad \text{birds: } y \propto x^{2.5} \]
Why: A solid object scaled up uniformly has volume, and so weight, proportional to the cube of any length. Birds come out at about 2.5 instead, meaning larger birds are proportionally lighter than pure geometric scaling predicts — which is what flight demands. Reading a fitted exponent against the exponent theory predicts is where these models stop being curve-fitting and start being science.
Ranking
From raw data to a power model.
Put in order
Why: Step four is the one with no counterpart in the semi-log case: the term with 2.5 times the logarithm of x has to be rewritten as the logarithm of x to the 2.5 before exponentiating can produce a power of x. That single application of Lesson 7.5's power property is what turns the line back into a curve.
Comparison
Fill the blanks. Everything follows from where x sits.
Comparison matrix
| Question | Exponential | Power |
|---|---|---|
| Form | y = ab^x | y = ax^b |
| Straightening plot | (x, ln y) | (ln x, ln y) |
| Slope means | ln b | b |
| Intercept means | ln a | ln a |
Both intercepts mean the same thing, so both need exponentiating. Only the slopes differ, and that difference is the whole reason for choosing one plot over the other.
Pattern
Two points, or a whole table.
Straightness of the transformed plot is not a formality. It is the test that decides whether the family fits at all.
OpenStax Algebra and Trigonometry 2e, §6.7 Exponential and Logarithmic Models §6.7
Check
Two points, exponential.
Check your understanding
Write y = ab^x through (1, 6) and (3, 24).
Answer: A
Why: Dividing gives b squared equal to 4, so b is 2 and a is 3.
Check
Which transformation straightens which data?
Check your understanding
To test whether data fits y = ax^b, what do you plot?
Answer: A
Why: A power model has the variable in the base, so both variables need the logarithm.
Check
Reading a log-log line.
Check your understanding
A log-log plot gives ln y = 2.5 ln x - 2.546. What is the model?
Answer: A
Why: The slope is the exponent, and e to the -2.546 is about 0.0784.
Real world
Kepler measured each planet's orbital period T in years and its distance x from the sun in astronomical units. For Earth both are 1; for Jupiter x is 5.20 and T is 11.86; for Saturn x is 9.54 and T is 29.4.
Discussion prompt
Plot the logarithms of both variables for these three planets, find the slope, and say what law you have rediscovered.
Hint: Take natural logarithms of both coordinates.
Answer:
\[ (\ln x, \ln T): (0, 0), \; (1.649, 2.473), \; (2.256, 3.381) \]
\[ m = \frac{3.381 - 0}{2.256 - 0} \approx 1.499 \approx \tfrac{3}{2} \]
The slope is three halves, so T equals x to the three halves, or T squared equals x cubed — Kepler's third law, found in 1618 after decades of work.
The log-log plot is what makes it visible in three points. On ordinary axes the data curves and the exponent is invisible; transformed, it is the slope of a line you can measure with a ruler. This is the same reading that gave the birds their 2.5, and it is why scientists reach for log-log paper whenever they suspect one quantity scales as a power of another.
Commit first
Answer, then rate your confidence honestly.
Predict first
Both y equals ab to the x and y equals ax to the b have two constants. Does that make them interchangeable?
Correct: No — one has the variable in the exponent and the other in the base, and they behave completely differently.
\[ y = ab^x: \; x \text{ in the exponent}; \qquad y = ax^b: \; x \text{ in the base} \]
Why: Sharing a constant count means they need the same number of points, nothing more. An exponential multiplies by a fixed factor for each unit increase in x; a power function does not. In the long run an exponential with any base above 1 overtakes any power function, however large its exponent. And they are detected by different transformations: the semi-log plot for one, the log-log plot for the other. Confusing them produces a model that fits the two chosen points and misses everything else.
Explain it
They can fit a line to a scatter plot and have just met logarithms.
Discussion prompt
In four sentences or fewer, explain why taking the logarithm of y turns exponential data into a straight line.
Hint: Take the logarithm of both sides of the model.
Answer:
Start from the model: y equals a times b to the x. Take the natural logarithm of both sides, and the product becomes a sum while the exponent comes down as a factor.
What is left is the logarithm of y equal to x times the logarithm of b, plus the logarithm of a — which is a line in x, with slope and intercept both constants. So plotting the logarithm of y against x must give a line whenever the original data is exponential.
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 the exponential case, divide the two equations and read the ratio as b raised to the difference of the x values. For the power case, do the same and finish with a logarithm. For choosing a plot, ask where the variable sits. For converting back, remember that a slope is the exponent on a log-log plot but the logarithm of the base on a semi-log one, and that every intercept has to be exponentiated.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a modelling page. Left column, exponential: write the form, work Example 1 through two points, then derive the transformed equation by taking the logarithm of both sides and label which part of it is the slope and which the intercept. Right column, power: write the form, work Example 4 through two points, then derive its transformed equation the same way and label the slope as the exponent. Across the bottom, sketch four small axes — an exponential curve, its semi-log line, a power curve, its log-log line — and under each write one sentence saying what is plotted. In the middle, write the one question that decides between the two families, and a worked example of the scooter data from table to model to prediction. Finish with a boxed sentence on why a good fit does not license a far prediction.
If your two derivations look identical, check the power one again: it needs a logarithm on x as well, which is exactly what makes its slope the exponent rather than the logarithm of a base.
Recap
Five things, and every one of them turns a curve into a line.
| If you see | Then |
|---|---|
| Two points and y = ab^x | Divide the equations; the ratio is b to the exponent difference |
| Two points and y = ax^b | Divide, then take a logarithm to free b |
| (x, ln y) nearly linear | An exponential model fits |
| (ln x, ln y) nearly linear | A power model fits |
| A semi-log slope | It is ln b, so exponentiate it |
| A log-log slope | It is b itself |
That completes Chapter 7. Chapter 8 turns to rational functions, where the variable moves into a denominator and a new kind of asymptote appears.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.7 Write and Apply Exponential and Power Functions §7.7, pp. 528-535 — everything on these slides traces back here
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