Question

Logarithmic and Exponential Forms Complete the table by finding the appropriate logarithmic or exponential form of the equation, as in Example 1.

Short answer

Logarithmic form	Exponential form
${\log }_4\left(64\right)=3$	$4^3=64$
${\log }_4\left(2\right)=\frac{1}{2}$	$4^{1/2}=2$
${\log }_4\left(8\right)=\frac{3}{2}$	$4^{3/2}=8$
${\log }_4\left(\frac{1}{16}\right)=-2$	$4^{-2}=\frac{1}{16}$
${\log }_4\left(\frac{1}{2}\right)=-\frac{1}{2}$	$4^{-1/2}=\frac{1}{2}$
${\log }_4\left(\frac{1}{32}\right)=-\frac{5}{2}$	$4^{-5/2}=\frac{1}{32}$

Step-by-step solution

Step 1. Given.

The given table is:

Step 2. To determine.

We have to fill in the blanks.

Step 3. Calculation.

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

So, the table will be:

Logarithmic form (${\log }_{a}c=b$)	a	b	c	Exponential form ($a^b=c$)
${\log }_4\left(64\right)=3$	4	3	64	$4^3=64$
${\log }_4\left(2\right)=\frac{1}{2}$	4	$\frac{1}{2}$	2	$4^{1/2}=2$
${\log }_4\left(8\right)=\frac{3}{2}$	4	$\frac{3}{2}$	8	$4^{3/2}=8$
${\log }_4\left(\frac{1}{16}\right)=-2$	4	$-2$	1/16	$4^{-2}=\frac{1}{16}$
${\log }_4\left(\frac{1}{2}\right)=-\frac{1}{2}$	4	$-\frac{1}{2}$	$\frac{1}{2}$	$4^{-1/2}=\frac{1}{2}$
${\log }_4\left(\frac{1}{32}\right)=-\frac{5}{2}$	4	$-\frac{5}{2}$	$\frac{1}{32}$	$4^{-5/2}=\frac{1}{32}$