Question

In the following exercises, solve for x.

${\log }_{2}x+{\log }_2\left(x-2\right)=3$

Short answer

The value of x for the equation${\log }_{2}x+{\log }_2\left(x-2\right)=3$ is$4$

Step-by-step solution

Step 1. Given data 

The given Logarithmic Equation is${\log }_{2}x+{\log }_2\left(x-2\right)=3$

Step 2. Apply the product properties of logarithm  

Use product property of logarithm${\log }_a\left(m·n\right)={\log }_a\left(m\right)+{\log }_a\left(n\right)$

$$\begin{gathered} {\log }_{2}x+{\log }_2\left(x-2\right)=3 \\ {\log }_2\left(x\left(x-2\right)\right)=3 \\ {\log }_2\left(x^2-2x\right)=3 \end{gathered}$$

Step 3. Write the equation in exponential form 

Change the logarithm equation into an exponential equation

$$\begin{gathered} {\log }_2\left(x^2-2x\right)=3 \\ {2}^3=x^2-2x \\ 8=x^2-2x \\ 0=x^2-2x-8 \\ 0=(x+2)(x-4) \end{gathered}$$

so $x=-2,4$

The logarithm of a negative number is not possible

so $x=4$

Step 4. Verification 

Substitute $4$for $x$in the equation

$$\begin{gathered} {\log }_{2}x+{\log }_2\left(x-2\right)=3 \\ {\log }_{2}4+{\log }_2\left(4-2\right)\stackrel{?}{=}3 \\ {\log }_{2}4+{\log }_{2}2\stackrel{?}{=}3 \\ {\log }_2(4·2)\stackrel{?}{=}3 \\ {\log }_2{2}^3\stackrel{?}{=}3 \\ 3=3 \end{gathered}$$