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
${\log }_8\left(8\right)=1$	$8^1=8$
${\log }_8\left(8\right)=64$	$8^2=64$
${\log }_8\left(4\right)=\frac{2}{3}$	$8^{2/3}=4$
${\log }_8\left(512\right)=3$	$8^3=512$
${\log }_8\left(\frac{1}{8}\right)=-1$	$8^{-1}=\frac{1}{8}$
${\log }_8\left(\frac{1}{64}\right)=-2$	$8^{-2}=\frac{1}{64}$

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 ${\log }_{a}c=b$.

In ${\log }_{a}c=b\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\log }_{8}8=1$, $a=8,b=1,c=8$. So, the exponential form is: $a^b=c\text{\hspace{0.17em}\hspace{0.17em}or},8^1=8$.

In ${\log }_{a}c=b\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\log }_{8}64=2$, $a=8,b=2,c=64$. So, the exponential form is: $a^b=c\text{\hspace{0.17em}\hspace{0.17em}or},8^2=64$.

In $a^b=c\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or},8^{2/3}=4$, $a=8,b=\frac{2}{3},c=4$. So, the logarithmic form is: ${\log }_{a}c=b\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\log }_8\left(4\right)=\frac{2}{3}$.

In $a^b=c\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or},8^3=512$, $a=8,b=3,c=512$. So, the logarithmic form is: ${\log }_{a}c=b\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\log }_8\left(512\right)=3$.

In ${\log }_{a}c=b\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\log }_8\left(\frac{1}{8}\right)=-1$, $a=8,b=-1,c=\frac{1}{8}$. So, the exponential form is: $a^b=c\text{\hspace{0.17em}\hspace{0.17em}or},8^{-1}=\frac{1}{8}$.

In $a^b=c\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or},8^{-2}=\frac{1}{64}$, $a=8,b=-2,c=\frac{1}{64}$. So, the logarithmic form is: ${\log }_{a}c=b\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\log }_8\left(\frac{1}{64}\right)=-2$.

So, the table is:

Logarithmic form	Exponential form
${\log }_8\left(8\right)=1$	$8^1=8$
${\log }_8\left(8\right)=64$	$8^2=64$
${\log }_8\left(4\right)=\frac{2}{3}$	$8^{2/3}=4$
${\log }_8\left(512\right)=3$	$8^3=512$
${\log }_8\left(\frac{1}{8}\right)=-1$	$8^{-1}=\frac{1}{8}$
${\log }_8\left(\frac{1}{64}\right)=-2$	$8^{-2}=\frac{1}{64}$