How Do You Solve an Exponential Equation Using Logarithms?

Take a logarithm of both sides, use the power law to bring the exponent down as a coefficient, and then isolate the unknown. For $a^{f(x)}=b$, this gives $f(x)=\dfrac{\ln b}{\ln a}$, provided $a>0$, $a\neq 1$, and $b>0$.

The Method

Suppose

$$a^{f(x)}=b.$$

Take natural logarithms of both sides:

$$\ln\left(a^{f(x)}\right)=\ln b.$$

Apply the power law of logarithms, $\ln(u^k)=k\ln u$:

$$f(x)\ln a=\ln b.$$

Therefore,

$$f(x)=\frac{\ln b}{\ln a}.$$

Any logarithm base can be used, but $\ln$ is usually most convenient. This method is needed when the two sides cannot easily be rewritten with the same base.

Worked Example

Solve $3^{2x-1}=7$.

Taking natural logarithms and using the power law:

$$(2x-1)\ln 3=\ln 7.$$

Hence,

$$x=\frac{1}{2}\left(1+\frac{\ln 7}{\ln 3}\right)\approx 1.39.$$

Always check that every logarithm has a positive argument. In the standard form above, $b>0$ is required because $\ln b$ is defined only for positive real numbers. Since $a^{f(x)}>0$ whenever , an equation such as has no real solution; taking is invalid.

Another useful pattern is $e^{3x+2}=11$. Since $\ln(e^u)=u$, taking $$ gives , so . This demonstrates that logarithmic and exponential functions are inverses. Substitute the exact result into the original equation during checking if algebraic rearrangement could have introduced an invalid answer.

A common misconception is that $\ln(a+b)=\ln a+\ln b$. This is false: logarithm laws apply to products, quotients, and powers, not sums.

IB Exam Technique

This is IB Mathematics: Analysis and Approaches, SL 1.7: Laws of exponents and logarithms and is also core content for AA HL. For Paper 1, leave an exact logarithmic answer where appropriate; on Paper 2, use your GDC and give $3$ significant figures unless instructed otherwise. For Find, show the logarithm step and power law rather than only the answer.