Precalculus · S.ID.B.6A

Building an exponential model from two points

An algae bloom covers 36 square meters on day 1 and 972 on day 4. If it multiplies by the same factor every day, what was that factor, and where did it start?

You need

Solving exponential equations by matching basesAlgebra 2 Growth and decay factorsAlgebra 1

Unlocks

Logistic growth: when growth levels offPrecalculus Semi-log plots: straightening exponential data with logarithmsPrecalculus

Where you are 6 in Exponential growth and decay

Functions13

Exponential functions6

Exponential growth and decay6

Doubling time and half life Growth and decay factors

Building an exponential model from two points

Linear versus exponential growth Logistic growth: when growth levels off Linear, polynomial and exponential growth compared

When you reach for it

A quantity changes by the same factor every period: an algae bloom that grows by the same percent each day, a concentration that falls by the same percent each day. Its rule is y = a \cdot b^x, with a starting value a and a factor b per period, as in Growth and decay factors. Often neither number is given. What you have instead is two measurements: the value at one time and the value at another.

Two unknowns need two facts, and two readings supply exactly two. This move turns two points (x_1, y_1) and (x_2, y_2) into the whole model. It is the exponential partner of finding a line from two points with The slope formula and the four kinds of slope: a line is fixed by the difference of the outputs, an exponential by their ratio.

The procedure

Two equations. Each point lies on the graph of y = a \cdot b^x, so each makes a true equation:

y_1 = a \cdot b^{x_1} \qquad y_2 = a \cdot b^{x_2}

Divide to remove a. Here a \neq 0, so the first equation can divide the second. The a cancels, and powers of the same base divide by subtracting exponents:

\frac{y_2}{y_1} = \frac{a \cdot b^{x_2}}{a \cdot b^{x_1}} = b^{x_2 - x_1}

One equation in one unknown is left. It says that over x_2 - x_1 periods the quantity is multiplied by \displaystyle \frac{y_2}{y_1}, which is the meaning of a factor.

Take a root to find b. Write n = x_2 - x_1. The factor is positive, and a positive number has exactly one positive nth root, so

b = \left(\frac{y_2}{y_1}\right)^{1/n}

(Rational exponents). When the ratio is a known power, such as 27 = 3^3, the root is exact: b^3 = 3^3 with both bases positive forces b = 3, the same reasoning as Solving exponential equations by matching bases with the roles of base and exponent swapped. Otherwise a calculator gives a decimal.

Back-substitute to find a. From y_1 = a \cdot b^{x_1}, dividing by b^{x_1} gives a = \dfrac{y_1}{b^{x_1}}. If one of the points is (0, y_1), there is nothing to do: b^0 = 1, so a = y_1.

Or skip a altogether. Since a \cdot b^{x_1} = y_1,

y = a \cdot b^{x} = a \cdot b^{x_1} \cdot b^{x - x_1} = y_1 \cdot b^{x - x_1}

This point form starts the clock at the first reading. It is the same function, and it is often the cleanest way to predict.

Check. Put both inputs back into the finished rule. Both outputs must come back.

Worked examples

Example 1

Two readings three days apart

Eustachy measures an algae bloom on a lake. It covers 36 square meters on day 1 and 972 square meters on day 4. Assuming the cover is multiplied by the same factor each day, write the cover A(t) on day t and say what each number in the rule means.

Solution

The points are (1, 36) and (4, 972), so 36 = a \cdot b and 972 = a \cdot b^4. Dividing the second equation by the first cancels a:

\frac{972}{36} = b^{4 - 1}, \qquad 27 = b^3

Since 27 = 3^3 and b is positive, b = 3. Then 36 = a \cdot 3 gives a = 12, so

A(t) = 12 \cdot 3^t

Check: A(1) = 36 and A(4) = 12 \cdot 81 = 972. The factor 3 says the cover triples every day, a growth of 200\% a day. The 12 is A(0): the model says the bloom covered 12 square meters on day 0. Nobody measured that; it comes from running the factor backward one day from 36.

Example 2

A decay through a negative input

Find the exponential function f(x) = a \cdot b^x whose graph passes through (-2, 50) and (1, 3.2).

Solution

The two equations are 50 = a \cdot b^{-2} and 3.2 = a \cdot b^1. Divide the later point’s equation by the earlier one’s, so the exponent gap is positive: 1 - (-2) = 3.

\frac{3.2}{50} = b^3, \qquad 0.064 = b^3

Since 0.4^3 = 0.064, b = 0.4. Then 3.2 = a \cdot 0.4 gives a = 8, and

f(x) = 8(0.4)^x

The factor is below 1, so this is decay: each step of 1 in x keeps 40\% of the value. Check the first point: \displaystyle f(-2) = 8 \cdot 0.4^{-2} = \frac{8}{0.16} = 50.

Example 3

Choosing where time starts

Aizhan keeps records of the lake’s bloom at its peak each summer: 12 acres in 2018 and 30 acres in 2023. Assuming the peak area grows by the same factor each year, write a model, give the yearly growth rate, and predict the peak area in 2026.

Solution

The year itself is a poor input: it would put b^{2018} in the equations. Let t count years since 2018 instead. Then the readings are (0, 12) and (5, 30), and the first one gives a = 12 at once.

The second gives 30 = 12 \cdot b^5, so b^5 = 2.5 and

b = 2.5^{1/5} \approx 1.2011

The model is A(t) \approx 12(1.2011)^t, a growth of about 20.1\% a year.

The year 2026 is t = 8. Using the exact factor,

A(8) = 12 \cdot \left(2.5^{1/5}\right)^8 = 12 \cdot 2.5^{8/5} \approx 51.99

so the model predicts a peak of about 52 acres. Keeping 2.5^{1/5} unrounded until the last step means the rounding happens once.

Example 4

Neither reading at the start

Bongani measures a bloom in another bay: 150 square meters on day 2 and 420 square meters on day 7. Write the model, and predict the cover on day 10.

Solution

The equations are 150 = a \cdot b^2 and 420 = a \cdot b^7. Dividing,

\frac{420}{150} = b^5, \qquad 2.8 = b^5

so b = 2.8^{1/5} \approx 1.2287: about 22.9\% growth a day. For a, divide the first equation by b^2:

a = \frac{150}{2.8^{2/5}} \approx \frac{150}{1.5096} \approx 99.36

So A(t) \approx 99.36(1.2287)^t. Figure 1 shows the curve through both readings.

Figure 1Example 4. Five days lie between the readings, and over those five days the cover is multiplied by \displaystyle \frac{420}{150} = 2.8.

For day 10, the point form is quicker. Day 10 is 3 days after the second reading, so

A(10) = 420 \cdot 2.8^{3/5} \approx 420 \cdot 1.8548 \approx 779.0

about 779 square meters. The full rule agrees: 99.36 \cdot 1.2287^{10} \approx 779.

Example 5

Point form, decay, and a threshold

After the lake is treated, Eustachy tracks the chlorophyll in the water, a measure of how much algae is present. It reads 48 micrograms per liter on day 3 and 15 micrograms per liter on day 10. Write the model in point form, find the daily percent decay, and find the first whole day the reading is below 5.

Solution

Start the clock at day 3: C(t) = 48 \cdot b^{t - 3}. Day 10 is 7 days later, so

15 = 48 \cdot b^7, \qquad b^7 = \frac{15}{48} = 0.3125

and b = 0.3125^{1/7} \approx 0.8469. Each day keeps about 84.7\% of the chlorophyll, a decay of about 15.3\% a day.

Seven more days multiply by 0.3125 again, so day 17 gives exactly

C(17) = 15 \cdot 0.3125 = 4.6875

which is below 5. One day earlier, \displaystyle C(16) = \frac{4.6875}{b} \approx 5.53, still above. So day 17 is the first whole day below 5. Finding the exact moment, not just the whole day, means solving 48 \cdot b^{t-3} = 5 for t, which takes logarithms (Solving by taking a logarithm of both sides).

Example 6

Do two points prove it is exponential?

Aizhan counts algae cells in a water sample, in thousands per milliliter: 60 on day 0 and 135 on day 2. Build an exponential model and a linear model through these two readings and compare their predictions for day 6. Her day-6 count is 680. Which model does it support?

Solution

Exponential: a = 60 and 60 \cdot b^2 = 135, so b^2 = 2.25 and b = 1.5. The model N(t) = 60(1.5)^t predicts N(6) = 60 \cdot 11.390625 \approx 683.4.

Linear: the slope is \displaystyle \frac{135 - 60}{2 - 0} = 37.5, so L(t) = 60 + 37.5t, which predicts L(6) = 285.

Figure 2Example 6. Two readings fit a line and an exponential equally well. A third reading tells them apart.

Both models pass through both readings, so the two readings cannot choose between them. The day-6 count of 680 sits within 4 of the exponential prediction and far from the linear one, so it supports the exponential model. It supports it; it does not prove it. The model is a claim that the factor stays the same, and only more data can test that claim.

Boundaries

The inputs must differ. If x_1 = x_2, dividing gives b^0 = 1, which says nothing about b. Two different outputs at the same input fit no function at all.

The outputs must be nonzero and have the same sign. For b > 0, b^x is always positive, so a \cdot b^x has the sign of a at every input. A zero output, or one positive and one negative output, fits no exponential model. Two negative outputs are fine: the ratio is positive and a comes out negative.

With those conditions, the answer is unique. The ratio \displaystyle \frac{y_2}{y_1} is positive and has exactly one positive nth root, so there is exactly one b, and then exactly one a. If the two outputs are equal, that b is 1 and the “model” is the constant function y = y_1: nothing grows or decays, so the data show no exponential change at all.

Rounded factors drift. A factor rounded to four decimal places is raised to a power when the model predicts, and the error grows with the power. Keep the root exact, or keep many places, until the last step.

Two points always fit. Any two readings meeting the conditions above fit an exponential model, and a linear one too. Whether the quantity really multiplies by a constant factor is a separate question that needs more data. A real bloom cannot grow by 23\% a day forever: the lake runs out of room and nutrients, and growth slows, which Logistic growth: when growth levels off describes. Predictions far outside the measured interval deserve the least trust.

Where this goes next

Taking the logarithm of y = a \cdot b^x gives \log y = \log a + x \log b, a straight line in x. On a semi-log plot the two readings become two points on a line, and this lesson becomes finding a slope and an intercept. With many readings rather than two, no single curve passes through them all, and Fitting linear and exponential models with technology finds the best one. Once b is known, Doubling time and half life asks how long it takes to double or halve.

Exercises

Exact factors

Find the exponential function y = a \cdot b^x through the two points.

(1, 10) and (3, 250)
(2, 48) and (5, 3072)
(0, 9) and (3, 72)
(1, 160) and (3, 10)
(-1, 15) and (1, 0.6)
(-3, 2) and (1, 162)

Factors that need a root

Find the exponential function through the two points, with a and b to the given places.

(0, 500) and (6, 1250), b to four decimal places.
(1, 40) and (4, 25), b to four and a to two decimal places.
(2, 9) and (10, 30), b to four and a to two decimal places.

Models from the lake

A bloom’s peak area was 20 acres in 2020 and 45 acres in 2024. With t in years since 2020, write a model, give the yearly growth rate, and predict the peak area in 2027.
Aizhan counts 1{,}800 cells per milliliter on day 2 and 48{,}600 on day 5. Write the model N(t) and predict the count on day 8.
After treatment, Bongani measures a bloom of 64 acres on day 2 and 27 acres on day 5. Find the daily factor and percent decay, the model’s area on day 0, and the area on day 8.
Chlorophyll reads 30 micrograms per liter on day 1 and 12 on day 6. Find the daily factor to four decimal places and the first whole day the reading is below 3. Show the values you compare.

Predicting without a

Write the exponential function through (3, 40) and (8, 90) in point form y = 40 \cdot b^{x - 3}, and find its value at x = 13 without a calculator.
An exponential function passes through (4, 50) and (6, 72). Find its values at x = 8 and at x = 5.

Exponential, linear or neither

Is there an exponential function through all three of (1, 24), (3, 54) and (5, 121.5)? Is there a linear one?
Is there an exponential function through all three of (0, 5), (2, 20) and (3, 35)? Is there a linear one?

Reasoning

Explain why no function y = a \cdot b^x with b > 0 passes through (1, 4) and (3, -9).
Bongani divides the first equation by the second instead, getting \displaystyle \frac{y_1}{y_2} = b^{x_1 - x_2}. Show that he gets the same b.
Show that for an exponential function through (x_1, y_1) and (x_2, y_2) with positive outputs, the value halfway between, at \displaystyle x = \frac{x_1 + x_2}{2}, is \displaystyle \sqrt{y_1 y_2}.
A bloom grows from 12 acres to 48 acres in 4 years. Eustachy says it grew 300\%, so 75\% a year. Explain what is wrong and find the true yearly rate, assuming a constant factor.

Answers

y = 2 \cdot 5^x. \displaystyle b^2 = \frac{250}{10} = 25, so b = 5, and \displaystyle a = \frac{10}{5} = 2.
y = 3 \cdot 4^x. \displaystyle b^3 = \frac{3072}{48} = 64, so b = 4, and \displaystyle a = \frac{48}{16} = 3.
y = 9 \cdot 2^x. a = 9, and \displaystyle b^3 = \frac{72}{9} = 8, so b = 2.
y = 640(0.25)^x. \displaystyle b^2 = \frac{10}{160} = \frac{1}{16}, so \displaystyle b = \frac{1}{4}, and \displaystyle a = \frac{160}{0.25} = 640.
y = 3(0.2)^x. \displaystyle b^2 = \frac{0.6}{15} = 0.04, so b = 0.2, and a = 15 \cdot 0.2 = 3, since a = y_1 \cdot b^{1} when x_1 = -1.
y = 54 \cdot 3^x. \displaystyle b^4 = \frac{162}{2} = 81, so b = 3, and \displaystyle a = \frac{162}{3} = 54.
y \approx 500(1.1650)^x. a = 500 and b^6 = 2.5, so b = 2.5^{1/6} \approx 1.1650.
y \approx 46.78(0.8550)^x. \displaystyle b^3 = \frac{25}{40} = 0.625, so b = 0.625^{1/3} \approx 0.8550, and \displaystyle a = \frac{40}{b} \approx 46.78.
y \approx 6.66(1.1624)^x. \displaystyle b^8 = \frac{30}{9} = \frac{10}{3}, so b \approx 1.1624, and \displaystyle a = \frac{9}{b^2} \approx 6.66.
A(t) \approx 20(1.2247)^t, about 22.5\% a year, and about 83 acres in 2027. \displaystyle b^4 = \frac{45}{20} = 2.25, so b = 2.25^{1/4} = 1.5^{1/2} \approx 1.2247; A(7) = 20 \cdot 1.5^{7/2} \approx 82.67.
N(t) = 200 \cdot 3^t, and N(8) = 1{,}312{,}200 cells per milliliter. \displaystyle b^3 = \frac{48600}{1800} = 27, so b = 3; \displaystyle a = \frac{1800}{9} = 200; 200 \cdot 6561 = 1{,}312{,}200.
Factor \displaystyle \frac{3}{4}, a 25\% decay a day; day 0: \displaystyle \frac{1024}{9} \approx 113.8 acres; day 8: \displaystyle \frac{729}{64} \approx 11.4 acres. \displaystyle b^3 = \frac{27}{64}, so \displaystyle b = \frac{3}{4}; \displaystyle a = \frac{64}{9/16}; day 8 is \displaystyle 27 \cdot \left(\frac{3}{4}\right)^3.
b = 0.4^{1/5} \approx 0.8326; first below 3 on day 14. \displaystyle b^5 = \frac{12}{30} = 0.4. With C(t) = 30 \cdot b^{t-1}, C(13) \approx 3.33 and C(14) \approx 2.77.
\displaystyle b^5 = \frac{90}{40} = 2.25, so y = 40 \cdot b^{x-3} with b = 2.25^{1/5}. At x = 13, x - 3 = 10 and b^{10} = 2.25^2 = 5.0625, so y = 202.5.
103.68 at x = 8 and 60 at x = 5. \displaystyle b^2 = \frac{72}{50} = 1.44, so b = 1.2; 72 \cdot 1.44 = 103.68 and 50 \cdot 1.2 = 60.
Exponential, yes: y = 16(1.5)^x. From the first two points b^2 = 2.25, b = 1.5, a = 16, and 16 \cdot 1.5^5 = 121.5, so the third point fits. Linear, no: the line through the first two has slope 15 and gives 84 at x = 5.
Neither. The exponential through the first two is y = 5 \cdot 2^x, which gives 40 at x = 3, not 35. The line through the first two is y = 5 + 7.5x, which gives 27.5 at x = 3.
For b > 0, b^x > 0 at every x, so a \cdot b^x has the same sign as a everywhere. A positive output at x = 1 forces a > 0, and a negative output at x = 3 forces a < 0. Both cannot hold. In the procedure, \displaystyle b^2 = -\frac{9}{4} has no real solution.
\displaystyle \frac{y_1}{y_2} = \left(\frac{y_2}{y_1}\right)^{-1} and x_1 - x_2 = -(x_2 - x_1), so by Powers of powers, products and quotients his root is

\left(\frac{y_1}{y_2}\right)^{\frac{1}{x_1 - x_2}} = \left(\frac{y_2}{y_1}\right)^{(-1) \cdot \frac{-1}{x_2 - x_1}} = \left(\frac{y_2}{y_1}\right)^{\frac{1}{x_2 - x_1}}

the same b. 20. In point form the value at the halfway input is y_1 \cdot b^{(x_2 - x_1)/2}. Since \displaystyle b^{x_2 - x_1} = \frac{y_2}{y_1}, this is \displaystyle y_1 \left(\frac{y_2}{y_1}\right)^{1/2} = \sqrt{y_1^2 \cdot \frac{y_2}{y_1}} = \sqrt{y_1 y_2}. 21. A percent per year multiplies; it does not add. At 75\% a year the bloom would be 12 \cdot 1.75^4 \approx 112.5 acres, not 48. The true factor solves \displaystyle b^4 = \frac{48}{12} = 4, so \displaystyle b = 4^{1/4} = \sqrt{2} \approx 1.4142, about 41.4\% a year.