How to evaluate logarithms by rewriting as exponents

A logarithm tells us which exponent produces a given number. Rewriting the logarithm as an exponential equation makes that exponent easier to find. Let’s use the same idea for whole numbers, fractional arguments, and bases less than 1.

Rewrite the logarithm in exponential form

The statement $\log_{b} (A) = x$ means $b^{x} = A$ . For real logarithms, the base satisfies $b > 0$ and $b \neq 1$ , and the argument satisfies $A > 0$ . For example, $\log_{2} (8) = 3$ because $2^{3} = 8$ .

\log_{b} (A) = x \Leftrightarrow b^{x} = A

Find a fractional exponent when the base is a power

For $\log_{27} (3)$ , let the answer be $x$ . Since $27 = 3^{3}$ , we can write both sides of the exponential equation with base $3$ .

\begin{array}{l} 27^{x} = 3 \\ (3^{3})^{x} = 3^{1} \\ 3^{3 x} = 3^{1} \\ 3 x = 1 \\ x = \frac{1}{3} \end{array}

Therefore, $\log_{27} (3) = 1 / 3$ . The exponent is fractional because the cube root of $27$ is $3$ .

Example: Evaluate a logarithm with base 243

Find $\log_{243} (3)$ . Rewrite $243$ as $3^{5}$ , then equate the exponents on the two sides.

\begin{array}{l} 243^{x} = 3 \\ (3^{5})^{x} = 3^{1} \\ 5 x = 1 \\ x = \frac{1}{5} \end{array}

Use a negative exponent for a reciprocal

To evaluate $\log_{2} (1 / 64)$ , express the fraction as a power of $2$ . The reciprocal $1 / 64$ is $2^{- 6}$ .

\begin{array}{l} 2^{x} = \frac{1}{64} \\ 2^{x} = \frac{1}{2^{6}} = 2^{- 6} \\ x = - 6 \end{array}

Example: Evaluate a logarithm with argument 1/625

Find $\log_{5} (1 / 625)$ . Since $625 = 5^{4}$ , the reciprocal is $5^{- 4}$ .

\begin{array}{l} 5^{x} = \frac{1}{625} \\ 5^{x} = 5^{- 4} \\ x = - 4 \end{array}

Example: Rewrite both numbers with a common base

Find $\log_{32} (16)$ . Both numbers are powers of $2$ : $32 = 2^{5}$ and $16 = 2^{4}$ .

\begin{array}{l} 32^{x} = 16 \\ (2^{5})^{x} = 2^{4} \\ 2^{5 x} = 2^{4} \\ 5 x = 4 \\ x = \frac{4}{5} \end{array}

Example: Evaluate a logarithm whose base is less than 1

Find $\log_{1 / 625} (5)$ . This time the base is a reciprocal, so rewrite the base as $5^{- 4}$ .

\begin{array}{l} (1 / 625)^{x} = 5 \\ (5^{- 4})^{x} = 5^{1} \\ - 4 x = 1 \\ x = - \frac{1}{4} \end{array}

In each example, changing from logarithmic form to exponential form gave us an equation for the unknown exponent. When both sides can be written with the same base, we can set their exponents equal.

Learn mathKrista KingMay 6, 2019math, learn online, online course, online math, algebra, algebra 2, algebra ii, logs, logarithms, evaluating logs, evaluating logarithms, solving logs, solving logarithms, general log rule, converting logs, logarithmic functions, exponential functions