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

Mar 9, 2016

You must use the log property ${\log}_{a} m + {\log}_{a} n = {\log}_{a} \left(m \times n\right)$

Note: ${\log}_{a} m - {\log}_{a} n = {\log}_{a} \left(\frac{m}{n}\right)$

Explanation:

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

Convert to exponential form.

${x}^{2} - 16 = {2}^{3}$

${x}^{2} - 24 = 0$

${x}^{2} = 24$

$x = \sqrt{24} = 2 \sqrt{6}$

Note: since the log of a negative number is non-defined, we cannot have $x = \pm \sqrt{24}$, we can only have $+ \sqrt{24}$

Practice exercises:

1. Solve for x.

a) ${\log}_{3} \left(2 x + 1\right) + {\log}_{3} \left(3 x - 4\right) = 6$

b) ${\log}_{4} \left(5 x - 2\right) - {\log}_{4} \left(3 x + 1\right) = 2$

Challenge problem:

Solve for x in the following equation.

$y = {\log}_{2} \left(4 x + 1\right) + {\log}_{2} \left(x + 3\right)$

Hopefully this helps!