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

Explanation:

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 $x=2\sqrt{6}$.