How do you solve Log_2 (x-4) + log_2 (x+4) = 3?

Dec 27, 2015

Use sum of the logarithms rule and condense the logarithms on the left. Covert logarithms to exponent form to solve. The steps are given below.

Explanation:

${\log}_{2} \left(x - 4\right) + {\log}_{2} \left(x + 4\right) = 3$

Let us use the rule $\log \left(A\right) + \log \left(B\right) = \log \left(A B\right)$

${\log}_{2} \left(x - 4\right) \left(x + 4\right) = 3$

${\log}_{2} \left({x}^{2} - 16\right) = 3$ Note : $\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$

Converting the logarithms to exponent form

If ${\log}_{b} \left(a\right) = k$ then $a = {b}^{k}$

${\log}_{2} \left({x}^{2} - 16\right) = 3$
${x}^{2} - 16 = {2}^{3}$
${x}^{2} - 16 = 8$
${x}^{2} - 16 + 16 = 8 + 16$
${x}^{2} = 24$
take square root on both the sides.
$x = \pm \sqrt{24}$
$x = \pm \sqrt{4 \cdot 6}$
$x = \pm \sqrt{4} \cdot \sqrt{6}$
$x = \pm 2 \sqrt{6}$

$x = 2 \sqrt{6} \mathmr{and} x = - 2 \sqrt{6}$

Check the validity of solution by substituting the solution in the original equation.

We can see for ${\log}_{2} \left(x - 4\right)$ if we substitute $x = - 2 \sqrt{6}$ the logarithm is not defined.

Therefore the final answer is $x = 2 \sqrt{6}$