Question

In the following exercises, solve for $x$

${\log }_5(x+1)+{\log }_5(x-5)={\log }_{5}7$

Short answer

By solving the equation ${\log }_5(x+1)+{\log }_5(x-5)={\log }_{5}7$, the value of $x$is $6$

Step-by-step solution

Step 1. Given

An expression${\log }_5(x+1)+{\log }_5(x-5)={\log }_{5}7$

To solve the expression for$x$

Step 2. Use the Product property

By Product Property of logarithm

${\log }_{a}M+{\log }_{a}N={\log }_{a}MN$

$$\begin{gathered} {\log }_5\left(\right(x+1\left)\right(x-5\left)\right)={\log }_{5}7 \\ {\log }_5(x^2-4x-5)={\log }_{5}7 \end{gathered}$$

Step 3. Use One to One property

If ${\log }_{a}M={\log }_{a}N$, then $M=N$

$$\begin{gathered} x^2-4x-5=7 \\ {x}^2-4x-12=0 \end{gathered}$$

Step 4. Solve the equation

$$\begin{gathered} x^2-4x-12=0 \\ (x-6)(x+2)=0 \\ x=-2,6 \end{gathered}$$

Logarithm for a negative number does not exist.

So,$x=6$

Step 5. Check the solution

Substitute $x=6$in the original equation and solve.

$$\begin{gathered} {\log }_5(6+1)+{\log }_5(6-5)={\log }_{5}7 \\ {\log }_5\left(7\right)+{\log }_5\left(1\right)={\log }_{5}7 \\ {\log }_{5}7={\log }_{5}7 \end{gathered}$$

Hence the solution is checked.