Every logarithm statement is an exponential statement in disguise, and switching between the two forms is the main tool for working with What a Logarithm Is.
Key idea
$$\log_b y = x \quad\Longleftrightarrow\quad b^x = y$$ The base of the log is the base of the power, and the log itself is the exponent. Use this to evaluate logs, and to solve equations where the unknown is the input, the base or the exponent.
Why it works
This is the definition of a logarithm, read in both directions. Because $b^x$ and $\log_b x$ are inverse functions, each form undoes the other: $b^{\log_b y} = y$ and $\log_b (b^x) = x$.
Worked examples
Worked example: Switching forms
Write $\log_5 125 = 3$ in exponential form, and $10^{-2} = 0.01$ in log form.
- Base 5, exponent 3: $5^3 = 125$.
- Base 10, exponent $-2$: $\log 0.01 = -2$.
Answer: $5^3 = 125$ and $\log 0.01 = -2$
Worked example: Solving for the input
Solve $\log_4 x = 3$.
- Exponential form: $x = 4^3 = 64$.
Answer: $x = 64$
Worked example: Solving for the base
Solve $\log_x 49 = 2$.
- $x^2 = 49$, so $x = \pm 7$.
- A base must be positive, so $x = 7$.
Answer: $x = 7$
Because the two forms undo each other, $b^{\log_b y} = y$ and $\log_b(b^x) = x$. For example $10^{\log 7} = 7$ and $\ln(e^{-2}) = -2$. These two facts are how you will solve exponential and log equations: apply the inverse to both sides.
Worked example: Solving for the exponent
Solve $10^x = 250$ exactly, and estimate it.
- Log form: $x = \log 250$.
- $100 < 250 < 1000$, so $x$ is between 2 and 3 (a calculator gives about $2.398$).
Answer: $x = \log 250 \approx 2.398$
Common mistake
Swapping the roles: writing $\log_5 125 = 3$ as $3^5 = 125$ or $125^3 = 5$. Keep the base as the base. A check: the base raised to the log must give the number inside the log.