AP Precalculus · Lesson 3

Exponential equations and semi-log plots

Know Know it

To solve an exponential equation, first isolate the power, then take a log of both sides: bx=cb^x = c gives x=log⁡bc=ln⁡cln⁡bx = \log_b c = \frac{\ln c}{\ln b}. If both sides can be written with the same base, set the exponents equal.

To solve a log equation, combine the logs into one, then rewrite it as a power: log⁡b(expression)=k\log_b(\text{expression}) = k means expression =bk= b^k. Then check every answer in the original equation, because a log of zero or a negative number is undefined.

A semi-log plot puts log⁡y\log y on the vertical axis (or uses a log scale). If y=abxy = ab^x, then log⁡10y=log⁡10a+xlog⁡10b\log_{10} y = \log_{10} a + x\log_{10} b. That is a line with slope log⁡10b\log_{10} b and vertical intercept log⁡10a\log_{10} a. So exponential data look linear on a semi-log plot.

Calculator On the exam

A graphing calculator is required on multiple-choice Part B and free-response Part A, in radian mode, and is not permitted on multiple-choice Part A or free-response Part B. Practice solving with logs by hand, and use the calculator for decimal values when it is allowed.

Trap Check for extraneous solutions

Solving log⁡3x+log⁡3(x+8)=2\log_3 x + \log_3(x + 8) = 2 gives x(x+8)=9x(x + 8) = 9, so x=1x = 1 or x=−9x = -9. Only x=1x = 1 works: log⁡3(−9)\log_3(-9) is undefined, so -9 is extraneous.

Worked example Easy

What value of x solves 5⋅2x=1605 \cdot 2^x = 160?

  1. 32
  2. 5Answer
  3. 16
  4. log⁡2155\log_2 155

How to solve it

  1. Isolate the power: 2x=160÷5=322^x = 160 \div 5 = 32.
  2. Write 32 as a power of 2: 32=2532 = 2^5.
  3. So x=5x = 5.

Why each choice is right or wrong

  • A. Incorrect. 32 is the value of 2x2^x, not of x; one more step gives x=5x = 5.
  • B. Correct. Divide by 5 to get 2x=322^x = 32, and 25=322^5 = 32, so x=5x = 5.
  • C. Incorrect. 16 is 160÷10160 \div 10, which treats 5⋅2x5 \cdot 2^x as 10x10x, a product instead of a power.
  • D. Incorrect. log⁡2155\log_2 155 subtracts 5 instead of dividing by it, but 5 multiplies the power.

Explain Explain it

Task verbs: justify, calculate

To justify that data are exponential, show that the log of y changes by a constant amount per unit of x, or that the semi-log plot is a line. Then undo the log: if log⁡10y=k+mx\log_{10} y = k + mx, then y=10k(10m)xy = 10^k(10^m)^x, so a=10ka = 10^k and b=10mb = 10^m.

Worked example Medium

What are all solutions of log⁡2x+log⁡2(x−3)=2\log_2 x + \log_2(x - 3) = 2?

  1. x=4x = 4 and x=−1x = -1
  2. x=−1x = -1 only
  3. x=4x = 4 onlyAnswer
  4. There are no solutions

How to solve it

  1. Combine: log⁡2(x(x−3))=2\log_2(x(x - 3)) = 2, so x2−3x=4x^2 - 3x = 4.
  2. Factor: (x−4)(x+1)=0(x - 4)(x + 1) = 0, so x=4x = 4 or x=−1x = -1.
  3. Check: -1 makes log⁡2x\log_2 x undefined, so only x=4x = 4 remains.

Why each choice is right or wrong

  • A. Incorrect. Both 4 and -1 solve the quadratic, but -1 makes log⁡2(−1)\log_2(-1) undefined, so it is extraneous.
  • B. Incorrect. -1 only keeps the extraneous root and drops 4, which does check: log⁡24+log⁡21=2\log_2 4 + \log_2 1 = 2.
  • C. Correct. x(x−3)=4x(x - 3) = 4 gives x=4x = 4 or x=−1x = -1, and only 4 keeps both logs defined.
  • D. Incorrect. No solutions is wrong because x=4x = 4 works: log⁡24+log⁡21=2+0=2\log_2 4 + \log_2 1 = 2 + 0 = 2.

Justify Your turn

Data plotted as log⁡10y\log_{10} y against x lie on the line log⁡10y=1+0.3x\log_{10} y = 1 + 0.3x. Justify that an exponential model fits the data, and work out its growth factor.

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Compare with a model answer

A full answer does this

  • Undoes the log to get y=10⋅(100.3)xy = 10 \cdot (10^{0.3})^x, an exponential form
  • Gives the growth factor 100.310^{0.3}, about 2

Model answer

Since log⁡10y=1+0.3x\log_{10} y = 1 + 0.3x, we get y=101+0.3x=10⋅(100.3)xy = 10^{1 + 0.3x} = 10 \cdot (10^{0.3})^x. That has the form abxab^x with a=10a = 10, so an exponential model fits. The growth factor is 100.3≈1.99510^{0.3} \approx 1.995, so y about doubles for each 1-unit increase in x.

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