# How do you solve log_2(9x+5)=2+log_2(x^2-1)?

Oct 16, 2016

$x = - \frac{3}{4} \mathmr{and} 3$

#### Explanation:

We need to write 2 in the form of log_2?
${2}^{2} = 4$ so ${\log}_{2} 4 = 2$

Rewrite the equation as
${\log}_{2} \left(9 x + 5\right) = {\log}_{2} 4 + {\log}_{2} \left({x}^{2} - 1\right)$
Or using the addition rule of logs
${\log}_{2} \left(9 x + 5\right) = {\log}_{2} 4 \left({x}^{2} - 1\right)$
Therefore
$9 x + 5 = 4 \left({x}^{2} - 1\right)$
Rearrange
$9 x + 5 = 4 {x}^{2} - 4$
$4 {x}^{2} - 9 x - 9 = 0$
Factorise
$\left(4 x + 3\right) \left(x - 3\right) = 0$
Thus $x = - \frac{3}{4} \mathmr{and} 3$