Question

Exponential Form Express the equation in exponential form.

(a) ${\log }_{3}5=x$ (b)${\log }_7\left(3y\right)=2$

Short answer

1. Exponential form of ${\log }_{3}5=x$is $3^x=5$.
2. Exponential form of ${\log }_7\left(3y\right)=2$is $7^2=3y$.

Step-by-step solution

Part a. Step 1. Given.

The given logarithmic form is ${\log }_{3}5=x$.

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.

Logarithmic form (${\log }_{a}c=b$)	a	b	c	Exponential form ($a^b=c$)
${\log }_{3}5=x$	3	x	5	$3^x=5$

Hence, exponential form of ${\log }_{3}5=x$is $3^x=5$.

Part b. Step 1. Given.

The given logarithmic form is ${\log }_7\left(3y\right)=2$.

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.

Logarithmic form (${\log }_{a}c=b$)	a	b	c	Exponential form ($a^b=c$)
${\log }_7\left(3y\right)=2$	7	2	3y	$7^2=3y$

Hence, exponential form of ${\log }_7\left(3y\right)=2$is $7^2=3y$.