Question

In the following exercise, solve for x.

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

Short answer

The solution of the given expression is$x=4$

Step-by-step solution

Step 1. Given information. 

The given expression is${\log }_{2}x+{\log }_2(x-3)=2$

Step 2.  Use logarithmic properties     

Using product rule, we get:

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

Writing it in exponential form, we get:

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

Step 3. Solve for x.    

Solving the equation, we get:

$$\begin{gathered} x^2-3x-4=0 \\ {x}^2-4x+x-4=0 \\ x\left(x-4\right)+\left(x-4\right)=0 \\ \left(x+1\right)\left(x-4\right)=0 \end{gathered}$$

$x=-1$or

$x=4$

We eliminate $x=-1$as the logarithmic function doesn't take any negative values.

Step 4. Check the value    

Substituting $x=4$in the given expression, we get:

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

This is true.

Thus, $x=4$is the solution of the given function.