Question

Logarithmic Form Express the equation in logarithmic form.

(a)$8^{-1}=\frac{1}{8}$ (b)$2^{-3}=\frac{1}{8}$

Short answer

1. Logarithmic form of $8^{-1}=\frac{1}{8}$is ${\log }_8\left(\frac{1}{8}\right)=-1$.
2. Logarithmic form of $2^{-3}=\frac{1}{8}$is ${\log }_2\left(\frac{1}{8}\right)=-3$.

Step-by-step solution

Part a. Step 1. Given.

The given exponential form is $8^{-1}=\frac{1}{8}$.

Part a. Step 2. To determine.

We have to find the exponential form.

Part a. Step 3. Calculation.

We’ll use the formula:$a^b=c$ is equivalent to ${\log }_{a}c=b$.

Comparing we find the values of a, b, c. Then plug them in the exponential form.

Exponential form ($a^b=c$)	a	b	c	Logarithmic form (${\log }_{a}c=b$)
$8^{-1}=\frac{1}{8}$	8	$-1$	$\frac{1}{8}$	${\log }_8\left(\frac{1}{8}\right)=-1$

Hence, logarithmic form of $8^{-1}=\frac{1}{8}$is ${\log }_8\left(\frac{1}{8}\right)=-1$.

Part b. Step 1. Given.

The given exponential form is $2^{-3}=\frac{1}{8}$.

Part b. Step 2. To determine.

We have to find the exponential form.

Part b. Step 3. Calculation.

We’ll use the formula:$a^b=c$ is equivalent to ${\log }_{a}c=b$.

Comparing we find the values of a, b, c. Then plug them in the exponential form.

Exponential form ($a^b=c$)	a	b	c	Logarithmic form (${\log }_{a}c=b$)
$2^{-3}=\frac{1}{8}$	2	$-3$	$\frac{1}{8}$	${\log }_2\left(\frac{1}{8}\right)=-3$

Hence, logarithmic form of $2^{-3}=\frac{1}{8}$is ${\log }_2\left(\frac{1}{8}\right)=-3$.