Chapter 8.2, Problem 13PPS

To determine

To calculate: The value of c in the equation $9^{3c+1}={27}^{3c-1}$ .

Answer to Problem 13PPS

The value of c in the equation $9^{3c+1}={27}^{3c-1}$ is $\frac{5}{3}$ .

Explanation of Solution

Given information:

The given equation is $9^{3c+1}={27}^{3c-1}$ .

Formula used:

Property of equality of exponential functions:

If $b^x=b^y$ then $x=y$for all $b>0$ ,and $b\ne 1$ .

Calculation:

Consider the given equation,

  $9^{3c+1}={27}^{3c-1}$

Since 9 can be written as $3^2$ and 27 can be written as $3^3$ .So, rewrite the above equation as like this,

  $\begin{array}{c}{9}^{3c+1}={27}^{3c-1}\\ 3^{2\left(3c+1\right)}=3^{3\left(3c-1\right)}\end{array}$

Now, apply the property of equality of exponential functions:

  $\begin{array}{c}2\left(3c+1\right)=3\left(3c-1\right)\\ 6c+2=9c-3\\ 6c-9c=-3-2\\ -3c=-5\end{array}$

Proceed to next,

  $\begin{array}{c}-3c=-5\\ c=\frac{-5}{-3}\\ c=\frac{5}{3}\end{array}$

Therefore, the value of c in the equation $9^{3c+1}={27}^{3c-1}$ is $\frac{5}{3}$ .