Chapter 5, Problem 19CT

To determine

The solution of the equation ${\log }_2\left(x-4\right)+{\log }_2\left(x+4\right)=3$ . Express irrational solutions in exact form.

Answer to Problem 19CT

Solution:

The solution of ${\log }_2\left(x-4\right)+{\log }_2\left(x+4\right)=3$ is $x=2\sqrt{6}$ .

Explanation of Solution

Given information:

The equation is ${\log }_2\left(x-4\right)+{\log }_2\left(x+4\right)=3$ .

Consider the equation ${\log }_2\left(x-4\right)+{\log }_2\left(x+4\right)=3$ .

The domain of the variable requires that $x-4>0$ and $x+4>0$ , so $x>4$ and $x>-4$ .

This means, any solution must satisfy $x>-4$ .

To obtain exact solution, first express left side as a single logarithm.

By using ${\log }_{a}M+{\log }_{a}N={\log }_a\left(MN\right)$ ,

  $\Rightarrow {\log }_2\left[\left(x-4\right)\left(x+4\right)\right]=3$

By using $y={\log }_{a}x$ if and only if $a^y=xa>0a\ne 1$ ,

  $\Rightarrow \left(x-4\right)\left(x+4\right)=2^3$

  $\Rightarrow x^2-16=8$

  $\Rightarrow x^2-24=0$

  $\Rightarrow x^2=24$

Taking square root of both sides,

  $\Rightarrow x=\pm \sqrt{24}$

  $\Rightarrow x=\pm 2\sqrt{6}$

  $\Rightarrow x=2\sqrt{6}$ or $x=-2\sqrt{6}$

Since $x=-2\sqrt{6}$ does not satisfy $x>-4$ ,

Hence, $x=2\sqrt{6}$ is the only solution of ${\log }_2\left(x-4\right)+{\log }_2\left(x+4\right)=3$ .

Therefore, the solution of ${\log }_2\left(x-4\right)+{\log }_2\left(x+4\right)=3$ is $x=2\sqrt{6}$ .