Question

In the following exercises, solve for $x$

${\log }_5(x+3)+{\log }_5(x-6)={\log }_{5}10$

Short answer

By solving the equation ${\log }_5(x+3)+{\log }_5(x-6)={\log }_{5}10$, the value of $x$is $7$

Step-by-step solution

Step 1. Given

An expression${\log }_5(x+3)+{\log }_5(x-6)={\log }_{5}10$

To solve the expression for$x$

Step 2. Use the Product property

By Product Property of logarithm,

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

$$\begin{gathered} {\log }_5\left(\right(x+3\left)\right(x-6\left)\right)={\log }_{5}10 \\ {\log }_5(x^2-3x-18)={\log }_{5}10 \end{gathered}$$

Step 3. Use One to One property

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

$$\begin{gathered} x^2-3x-18=10 \\ {x}^2-3x-28=0 \end{gathered}$$

Step 4. Solve the equation

$$\begin{gathered} x^2-3x-28=0 \\ (x-7)(x+4)=0 \\ x=-4,7 \end{gathered}$$

Logarithm for a negative number does not exist.

So,$x=7$

Step 5. Check the solution

Substitute $x=7$in the original equation and solve.

$$\begin{gathered} {\log }_5(7+3)+{\log }_5(7-6)={\log }_{5}10 \\ {\log }_5\left(10\right)+{\log }_5\left(1\right)={\log }_{5}10 \\ 10=10 \end{gathered}$$

Hence the solution is checked.