# How do you solve log_4 (x^2 + 3x) - log_4(x - 5) =1?

Jul 28, 2016

$x = \frac{1 + \sqrt{79} i}{2} , \frac{1 - \sqrt{79} i}{2}$

#### Explanation:

${\log}_{4} \left({x}^{2} + 3 x\right) - {\log}_{4} \left(x - 5\right) = 1$
Using the identity ${\log}_{a} b - {\log}_{a} c = {\log}_{a} \left(\frac{b}{c}\right)$, we can simplify the expression to
${\log}_{4} \left(\frac{{x}^{2} + 3 x}{x - 5}\right) = 1$
This yields to
$\frac{{x}^{2} + 3 x}{x - 5} = {4}^{1}$
This is because, if ${\log}_{a} b = c$, then $b = {a}^{c}$
Now the equation can be solved for values of $x$
$\implies {x}^{2} + 3 x = {4}^{1} \left(x - 5\right)$
$\implies {x}^{2} + 3 x = 4 x - 20$
$\implies {x}^{2} + 3 x - 4 x + 20 = 0$
$\implies {x}^{2} - x + 20 = 0$
The roots of this equation are imaginary.
They can be found using the formula $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
Here, $a = 1$, $b = - 1$ and $c = 20$
Therefore, $x = \frac{1 \pm \sqrt{{\left(- 1\right)}^{2} - 4 \cdot 1 \cdot 20}}{2 \cdot 1}$
$\implies x = \frac{1 + \sqrt{79} i}{2} , \frac{1 - \sqrt{79} i}{2}$