Question

Logarithmic Form Express the equation in logarithmic form.

(a)$6^2=36$ (b)${10}^{-1}=\frac{1}{10}$

Short answer

1. Logarithmic form of $6^2=36$is ${\log }_6\left(36\right)=2$.
2. Logarithmic form of ${10}^{-1}=\frac{1}{10}$is ${\log }_{10}\left(\frac{1}{10}\right)=-1$.

Step-by-step solution

Part a. Step 1. Given.

The given exponential form is $6^2=36$.

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 (${\log }_{a}c=b$)
$6^2=36$	6	2	36	${\log }_6\left(36\right)=2$

Hence, logarithmic form of $6^2=36$is ${\log }_6\left(36\right)=2$.

Part b. Step 1. Given.

The given exponential form is ${10}^{-1}=\frac{1}{10}$.

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 (${\log }_{a}c=b$)
${10}^{-1}=\frac{1}{10}$	10	$-1$	$\frac{1}{10}$	${\log }_{10}\left(\frac{1}{10}\right)=-1$

Hence, logarithmic form of ${10}^{-1}=\frac{1}{10}$is ${\log }_{10}\left(\frac{1}{10}\right)=-1$.