Chapter 8, Problem 47SGR

To determine

To solve: The equation ${\log }_{4}n+{\log }_4\left(n-4\right)={\log }_{4}5$ and check the solution.

Answer to Problem 47SGR

The exponential for of equation ${\log }_{4}n+{\log }_4\left(n-4\right)={\log }_{4}5$ is $n=5$ .

Explanation of Solution

Given:

The equation is ${\log }_{4}n+{\log }_4\left(n-4\right)={\log }_{4}5$ .

Calculation:

Simplify the equation ${\log }_{4}n+{\log }_4\left(n-4\right)={\log }_{4}5$ by using Product Rule.

  $\begin{array}{c}{\log }_{4}n+{\log }_4\left(n-4\right)={\log }_{4}5\\ {\log }_4\left[n\left(n-4\right)\right]={\log }_{4}5\\ n\left(n-4\right)=5\\ n^2-4n-5=0\end{array}$

Simplify further.

  $\begin{array}{c}{n}^2-5n+n-5=0\\ n\left(n-5\right)+1\left(n-5\right)=0\\ \left(n-5\right)\left(n+1\right)=0\\ n=5,-1\end{array}$

Take positive value.

Substitute $n=5$ in the equation

  ${\log }_{4}n+{\log }_4\left(n-4\right)={\log }_{4}5$ to check the solution.

  $\begin{array}{c}{\log }_{4}5+{\log }_4\left(5-4\right)={\log }_{4}5\\ {\log }_{4}5+{\log }_{4}1={\log }_{4}5\\ {\log }_{4}5+0={\log }_{4}5\\ {\log }_{4}5={\log }_{4}5\end{array}$

The left hand side is equal to the right hand side. So, the solution is correct.

Therefore, the solution of the equation ${\log }_{4}n+{\log }_4\left(n-4\right)={\log }_{4}5$ is $n=5$ .