Question

Solve: ${\log }_{2}x+{\log }_2(x-6)=4$

Short answer

By solving the equation, ${\log }_{2}x+{\log }_2(x-6)=4$, the value of $x$is $8$

Step-by-step solution

Step 1. Given

An expression${\log }_{2}x+{\log }_2(x-6)=4$

To solve the expression

Step 2. Use the product property on LHS

Use the Product Property of logarithm

$$\begin{gathered} {\log }_{a}M+{\log }_{a}N={\log }_a\left(MN\right) \\ {\log }_{2}x+{\log }_2(x-6)=4 \\ {\log }_2\left(x\right(x-6\left)\right)=4 \end{gathered}$$

Step 3. Rewrite in exponential form.

$$\begin{gathered} {\log }_2\left(x\right(x-6\left)\right)=4 \\ {x}^2-6x=2^4 \\ {x}^2-6x-16=0 \end{gathered}$$

Step 4. Factorize the equation

$$\begin{gathered} x^2-6x-16=0 \\ (x-8)(x+2)=0 \\ x=-2,8 \end{gathered}$$

Logarithm for a negative number does not exist.

So, $x=8$

Step 5. Check the solution

Substitute $x=8$in the original equation and solve.

$$\begin{gathered} {\log }_{2}8+{\log }_2(8-6)=4 \\ {\log }_2\left(8\right)+{\log }_2\left(2\right)=4 \\ {\log }_{2}16=4 \\ 4=4 \end{gathered}$$

Hence the solution has been checked.