Chapter 8.5, Problem 47PPS

To determine

To find: The value of each of the equation.

Answer to Problem 47PPS

The solution of the equation exists for the value $y=5$ .

Explanation of Solution

Given:

The given equation is ${\log }_4\left(2y+2\right)-{\log }_4\left(y-2\right)=1$ .

Calculation:

Consider the given equation is,

  ${\log }_4\left(2y+2\right)-{\log }_4\left(y-2\right)=1$

Solve the above equation as,

  $\begin{array}{l}{\log }_4\left(2y+2\right)-{\log }_4\left(y-2\right)=1\\ {\log }_4\frac{\left(2y+2\right)}{\left(y-2\right)}=1\\ \frac{\left(2y+2\right)}{\left(y-2\right)}=4^1\\ 2y+2=4\left(y-2\right)\end{array}$

Solve further as,

  $\begin{array}{l}2y=10\\ y=5\end{array}$

Thus, the solution of the equation exists for the value $y=5$ .