Solve Exponential Equation: Step-by-Step Guide
An exponential equation has the variable in the exponent, like 3^x = 27. If you can write both sides with the same base, then the exponents must be equal. If not, take a logarithm of both sides, and use the power rule to bring the exponent down so you can solve for x.
Math Racer skill guide · Algebra 2 · Grade 11
How to solve it, step by step
- Get the exponential part by itself. Use algebra to get the term with the variable in the exponent alone on one side.\[b^{f(x)} = c\]
- Match the bases or take a log. If both sides can be written with the same base, set the exponents equal. If not, take the log (or ln) of both sides.\[\ln\left(b^{f(x)}\right) = \ln(c)\]
- Bring the exponent down. Use the power rule: log(b^k) = k log(b). This moves the exponent to the front.\[f(x)\,\ln(b) = \ln(c)\]
- Solve for x. Divide by log(b) (or ln(b)), then solve the remaining equation.\[f(x) = \dfrac{\ln(c)}{\ln(b)}\]
- Check. Put your answer into the original equation, or check it with a calculator.\[b^{f(x_0)} \stackrel{?}{=} c\]
Worked example
Solve 3^(2x - 1) = 27.
Step 1: The exponential part is already by itself: 3^(2x - 1) = 27.
Step 2: Write 27 with base 3: 27 = 3^3. So 3^(2x - 1) = 3^3.
Step 3: The bases are the same, so set the exponents equal: 2x - 1 = 3.
Step 4: Solve: 2x = 4, so x = 2.
Check: 3^(2(2) - 1) = 3^3 = 27. Correct.
Another way (logs): take ln of both sides: (2x - 1) ln 3 = ln 27, so 2x - 1 = ln 27 / ln 3 = 3, and x = 2.
Tips and common mistakes
- First try to write both sides with the same base. For example, 8 = 2^3, 27 = 3^3, and 1/4 = 2^-2.
- When you take a log, you can use any base. ln and log base 10 are the most common.
- Use the power rule only AFTER you take the log of both sides.
- Be careful with negative exponents. b^-x = 1/b^x.
- If the equation looks like a x b^x = c, divide by a first so b^x is alone before you take the log.
- For e^x = c, take ln of both sides to get x = ln(c).
- The answer may not be a neat number. An exact answer like x = ln(7) / ln(3) is fine.
- Some equations hide a quadratic. For b^(2x), let u = b^x and solve for u first.