Precalculus Exponential and Logarithmic Equations Worksheet Solutions (PDF)
This exp log equations answers for Precalculus covers exponential and logarithmic equations (Common Core F-LE.A.4, F-BF.B.5, A-REI.D.11), with a free PDF to print.
Precalculus · Common Core State Standards: F-LE.A.4, F-BF.B.5, A-REI.D.11
Exp log equations answers: exercises 1 to 10
Exercise 1 – Same base
\(64=2^{6}\).
\(x=6\)
Exercise 2 – Same base with a shift
\(125=5^{3}\), so \(x-1=3\).
\(x=4\)
Exercise 3 – Rewrite the base
\(9^{x}=3^{2x}\) and \(27=3^{3}\), so \(2x=3\).
\(x=\frac32\)
Exercise 4 – Take a log
\(x\ln3=\ln20\), so \(x=\frac{\ln20}{\ln3}\approx 2.727\).
\(x\approx 2.727\)
Exercise 5 – Isolate first
Divide by 4: \(3^{x}=25\). Then \(x=\frac{\ln25}{\ln3}\approx 2.930\).
\(x\approx 2.930\)
Exercise 6 – Base e
\(2x=\ln15\), so \(x=\frac{\ln15}{2}\approx 1.354\).
\(x\approx 1.354\)
Exercise 7 – Base e with isolation
\(2e^{3x}=10\), \(e^{3x}=5\), \(3x=\ln5\), so \(x=\frac{\ln5}{3}\approx 0.536\).
\(x\approx 0.536\)
Exercise 8 – Simple log equation
\(x+3=2^{4}=16\).
\(x=13\)
Exercise 9 – Natural log equation
\(x-1=e^{3}\approx20.09\), so \(x=1+e^{3}\approx 21.09\).
\(x\approx 21.09\)
Exercise 10 – Combine logs
\(\log\left(x(x-3)\right)=1\), so \(x^{2}-3x=10\) and \((x-5)(x+2)=0\). \(x=-2\) makes \(\log(-2)\) undefined.
\(x=5\)
Exercises 11 and up
Exercise 11 – Combine base-2 logs
\(x(x-2)=2^{3}=8\), so \(x^{2}-2x-8=0\) and \((x-4)(x+2)=0\). Reject \(x=-2\).
\(x=4\)
Exercise 12 – Difference of logs
\(\frac{x+6}{x-1}=2^{3}=8\), so \(x+6=8x-8\) and \(7x=14\), \(x=2\). Check: both arguments are positive.
\(x=2\)
Exercise 13 – Solve by graphing
The curve meets the line near \(x=3.3\). Exactly, \(x=\frac{\ln10}{\ln2}\approx 3.32\).
\(x\approx 3.32\)
Exercise 14 – Savings goal
\(4000(1.05)^{t}=6000\), so \(1.05^{t}=1.5\) and \(t=\frac{\ln1.5}{\ln1.05}\approx 8.31\).
\(t\approx 8.31\) years
Exercise 15 – Doubling time
\(e^{0.06t}=2\), so \(t=\frac{\ln2}{0.06}\approx 11.55\).
\(11.55\) years
Exercise 16 – Half-life
\(50\) g: \(\left(\frac12\right)^{t/12}=\frac14\), so \(\frac{t}{12}=2\) and \(t=24\). \(30\) g: \(\left(\frac12\right)^{t/12}=0.15\), so \(t=12\cdot\frac{\ln(200/30)}{\ln2}\approx 32.84\).
\(24\) days; \(32.84\) days
Exercise 17 – Population growth
\(e^{0.04t}=3\), so \(t=\frac{\ln3}{0.04}\approx 27.47\).
\(27.47\) years
Exercise 18 – Hydrogen concentration
\([\text{H}^{+}]=10^{-5.3}\approx 5.01\times10^{-6}\).
\(\approx 5.01\times10^{-6}\)
Exercise 19 – Sound intensity
\(8.5=\log\frac{I}{I_{0}}\), so \(\frac{I}{I_{0}}=10^{8.5}\approx 3.16\times10^{8}\).
\(\approx 3.16\times10^{8}\)
Exercise 20 – Different bases
\((x+1)\ln2=x\ln3\), so \(x\ln2+\ln2=x\ln3\) and \(\ln2=x(\ln3-\ln2)\). Then \(x=\frac{\ln2}{\ln3-\ln2}=\frac{\ln2}{\ln1.5}\approx 1.710\).
\(x\approx 1.710\)
Key facts behind this exp log equations answers
- Same base: if b^m = b^n then m = n.
- Different bases: take ln of both sides, then x ln b = ln c.
- log_b u = c is equivalent to u = b^c.
- Check log solutions: arguments must be positive; reject extraneous values.
- Continuous doubling time is ln 2 divided by the rate r.
How to use this exp log equations answers
Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.
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