Unit 11 · Lesson 4 of 5
Solving exponential equations
You'll be able to: solve equations like \(3^{x+1} = 81\) and \(5 \cdot 2^x = 35\), with exact answers or rounded decimals, including doubling-time problems.
Step 1 of 5Warm-up
Two quick questions from earlier lessons. Answer from memory first, then check.
Find the exact value: \(\log_2 12 - \log_2 3\).
Quotient rule: \(\log_2 \dfrac{12}{3} = \log_2 4\). Since \(2^2 = 4\), the answer is \(2\).
Solve \(3x - 4 = x + 6\).
Subtract \(x\): \(2x - 4 = 6\). Add 4: \(2x = 10\). Divide by 2: \(x = 5\).
Step 2 of 5The idea
In an exponential equation the variable is stuck in the exponent, like \(2^x = 11\). There are two ways to get it out.
Way 1: same base. If both sides can be written as powers of the same base, the exponents must be equal: \(b^M = b^N\) means \(M = N\).
Way 2: take a log. Get the power alone on one side. Take \(\ln\) of both sides. The power rule brings the exponent down: \(\ln\left(b^x\right) = x\ln b\). Then divide.
Exact or decimal. An exact answer keeps the logs, like \(x = \dfrac{\ln 7}{\ln 2}\). For a decimal, use a calculator and round only at the end.
Way 2 always works. Way 1 is faster when you spot it: 8, 16, 32 are powers of 2; 9, 27, 81 are powers of 3; 25, 125 are powers of 5.
Growth problems use the same steps. For money growing continuously, \(A = Pe^{rt}\), the doubling time is the \(t\) that makes \(A = 2P\).
Why this shows up on the placement test: solving exponential equations is one of the listed MATH 1010 skills, often as a growth or interest question.
Step 3 of 5Worked examples
Example 1 (same base): \(3^{x+1} = 81\), then \(4^x = 8\)
- \(81 = 3^4\), so \(3^{x+1} = 3^4\). Same base on both sides.
- Set the exponents equal: \(x + 1 = 4\), so \(x = 3\). Check: \(3^{3+1} = 3^4 = 81\).
- For \(4^x = 8\), use base 2: \(4 = 2^2\) and \(8 = 2^3\). So \(\left(2^2\right)^x = 2^3\), which is \(2^{2x} = 2^3\).
- \(2x = 3\), so \(x = \dfrac32\). Power of a power: multiply the exponents.
Example 2 (take a log): \(5 \cdot 2^x = 35\)
- Get the power alone: divide by 5, so \(2^x = 7\). 7 is not a power of 2, so Way 1 won't work.
- Take ln of both sides: \(\ln\left(2^x\right) = \ln 7\).
- Power rule: \(x \ln 2 = \ln 7\). Divide by \(\ln 2\): \(x = \dfrac{\ln 7}{\ln 2}\). This is the exact answer.
- Calculator: \(x \approx 2.807\), so about \(2.81\). Check: \(5 \cdot 2^{2.81} \approx 35\).
Step 4 of 5Practice
Type your answer and press Check. If you miss twice, open the worked solution. Then mark how it went. Use a calculator only where the problem says so.
Solve \(2^x = 32\).
\(32 = 2^5\), so \(2^x = 2^5\) and \(x = 5\).
Solve \(3^{2x - 1} = 27\).
\(27 = 3^3\), so \(3^{2x-1} = 3^3\) and \(2x - 1 = 3\).
Add 1: \(2x = 4\). Divide by 2: \(x = 2\).
Solve \(9^x = 27\).
Both are powers of 3: \(9 = 3^2\) and \(27 = 3^3\). So \(\left(3^2\right)^x = 3^3\), which is \(3^{2x} = 3^3\).
\(2x = 3\), so \(x = \dfrac32\).
Solve \(5^{x+2} = \dfrac{1}{25}\).
\(\dfrac{1}{25} = \dfrac{1}{5^2} = 5^{-2}\), so \(5^{x+2} = 5^{-2}\).
\(x + 2 = -2\), so \(x = -4\).
Solve \(2^x = 11\). Give the exact answer (no decimals).
11 is not a power of 2, so take ln of both sides: \(\ln\left(2^x\right) = \ln 11\).
Power rule: \(x\ln 2 = \ln 11\). Divide: \(x = \dfrac{\ln 11}{\ln 2}\). (About 3.46. Makes sense: \(2^3 = 8\) and \(2^4 = 16\).)
Calculator OK. Solve \(4 \cdot 3^x = 50\). Round to the nearest hundredth.
Get the power alone: divide by 4, so \(3^x = 12.5\).
Take ln: \(x\ln 3 = \ln 12.5\), so \(x = \dfrac{\ln 12.5}{\ln 3} \approx \dfrac{2.5257}{1.0986} \approx 2.299\).
Rounded: \(x \approx 2.30\).
Calculator OK. $1000 is invested at 6% compounded continuously, \(A = 1000e^{0.06t}\). How many years until it doubles to $2000? Round to the nearest tenth of a year.
Set \(A = 2000\): \(1000e^{0.06t} = 2000\). Divide by 1000: \(e^{0.06t} = 2\).
Take ln: \(0.06t = \ln 2\), so \(t = \dfrac{\ln 2}{0.06} \approx \dfrac{0.6931}{0.06} \approx 11.55\).
Rounded: about \(11.6\) years.
A bacteria culture starts with 200 cells and doubles every 3 hours, so \(N = 200 \cdot 2^{t/3}\) after \(t\) hours. After how many hours are there 3200 cells? (No calculator needed.)
\(200 \cdot 2^{t/3} = 3200\). Divide by 200: \(2^{t/3} = 16\).
\(16 = 2^4\), so \(\dfrac{t}{3} = 4\) and \(t = 12\). Answer: 12 hours.
Step 5 of 5Wrap-up
You can now solve an exponential equation two ways: match the bases and set the exponents equal, or isolate the power, take ln of both sides, and divide.
Read or watch next: OpenStax Intermediate Algebra 2e, 10.5 Solve Exponential and Logarithmic Equations (free textbook, with more practice at the end of the section).
Also good: Khan Academy: Solving exponential equations using properties of exponents (short videos). Quick rules: Exponentials and logs cheat sheet.
Something unclear? Ask your teacher bot a follow-up question, for example: "How do I know when to match bases and when to take a log?" It knows this course and can explain things another way.