Question

In the following exercise, solve for x.

${\log }_{4}x+{\log }_{4}x=3$

Short answer

The solution of the given expression is$x=9$

Step-by-step solution

Step 1. Given information.

The given expression is${\log }_{4}x+{\log }_{4}x=3$

Step 2. Use logarithmic properties      

Using product rule, we get:

$$\begin{gathered} {\log }_{4}x+{\log }_{4}x=3 \\ {\log }_4\left(x·x\right)=3 \\ {\log }_4{x}^2=3 \end{gathered}$$

Writing it in exponential form, we get:

$$\begin{gathered} x^2=3^4 \\ {x}^2=81 \end{gathered}$$

Step 3. Solve for x.   

Solving the equation, we get:

$$\begin{gathered} x^2=81 \\ x=\pm \sqrt{81} \\ x=\pm 9 \end{gathered}$$

We eliminate $x=-9$as the logarithmic function doesn't take any negative values.

Step 4. Check the value   

Substituting $x=9$in the given expression, we get:

$$\begin{gathered} {\log }_{4}x+{\log }_{4}x=3 \\ {\log }_{4}9+{\log }_{4}9=3 \\ {\log }_4\left(9·9\right)=3 \\ {\log }_{4}81=2 \\ 3=3 \end{gathered}$$

This is true.

Thus, $x=9$is the solution of the given function.