Solving Exponential & Logarithmic Equations
Solving Exponential & Logarithmic Equations
Solving exponential and logarithmic equations brings together everything you have learned in this module. Exponential equations are solved by taking the logarithm of both sides, while logarithmic equations are solved by rewriting them in exponential form.
Many problems require you to apply the properties of logarithms to simplify the equation before solving. This lesson covers both types and emphasizes checking for extraneous solutions, which can arise when the argument of a logarithm becomes negative or zero.
Type 1: Same-Base Exponential Equations
If $b^m = b^n$ then $m = n$ (one-to-one property).
Solve $2^{3x-1} = 16$.
Write $16 = 2^4$: $\;2^{3x-1} = 2^4$.
$3x - 1 = 4 \implies x = \dfrac{5}{3}$.
Type 2: Using Logarithms
If the bases cannot be made equal, take the logarithm of both sides.
Solve $5^x = 20$.
$\ln(5^x) = \ln(20)$
$x \ln(5) = \ln(20)$
$$x = \frac{\ln(20)}{\ln(5)} \approx \frac{2.996}{1.609} \approx 1.861$$
Solve $3^{2x+1} = 7^{x-2}$.
Take $\ln$ of both sides:
$(2x+1)\ln 3 = (x-2)\ln 7$
$2x\ln 3 + \ln 3 = x\ln 7 - 2\ln 7$
$x(2\ln 3 - \ln 7) = -2\ln 7 - \ln 3$
$$x = \frac{-2\ln 7 - \ln 3}{2\ln 3 - \ln 7} \approx \frac{-4.990}{0.249} \approx -20.04$$
Type 3: Logarithmic Equations
Solve $\log_2(x-3) = 4$.
Convert to exponential: $x - 3 = 2^4 = 16 \implies x = 19$.
Check: $\log_2(16) = 4$ ✓
Solve $\log(x) + \log(x+3) = 1$.
Condense: $\log[x(x+3)] = 1$.
$x(x+3) = 10^1 = 10$.
$x^2 + 3x - 10 = 0 \implies (x+5)(x-2) = 0$.
$x = -5$ or $x = 2$.
Check $x=-5$: $\log(-5)$ is undefined ✗
Check $x=2$: $\log(2) + \log(5) = \log(10) = 1$ ✓
Solution: $x = 2$.
Solve $\ln(2x+1) = 3$.
$2x + 1 = e^3 \implies x = \dfrac{e^3 - 1}{2} \approx 9.543$
Practice Problems
Show Answer Key
1. $x = 3$
2. $x = \frac{\ln 50}{\ln 3} \approx 3.561$
3. $x = 2$ ($2^5 = 32$)
4. $x = \frac{\ln 15}{2} \approx 1.354$
5. $x = 81$
6. $x = 10$ ($x = -1$ is extraneous)
7. $x = 4$ ($5^3 = 125$)
8. $x = e^2 \approx 7.389$
9. $x = \frac{5\ln 3}{2\ln 7 - \ln 3} \approx 2.087$
10. $x = 8$ ($x = -2$ is extraneous)
11. $x = 1 + \log(500) \approx 3.699$
12. $t = \frac{\ln 2}{0.06} \approx 11.55$ years
13. $x = 8$
14. $x = e^4 \approx 54.598$
15. $x = \frac{3}{2}$ ($9 = 3^2$, $27 = 3^3$, so $3^{2x} = 3^3$)