# How do you solve (log_2(2x-3) / log_2(x)) - log_x(x+6) + (1/log_ (x+2)(x)) = 1?

Mar 6, 2016

${x}^{3} + 8 {x}^{2} + 10 x + 3 = 0$

#### Explanation:

$\left({\log}_{2} \frac{2 x - 3}{\log} _ 2 x\right) - {\log}_{x} \left(x + 6\right) + \left(\frac{1}{\log} _ \left(x + 2\right) x\right) = 1$
${\log}_{x} \left(x + 6\right) = \frac{{\log}_{2} \left(x + 6\right)}{{\log}_{2} x}$
${\log}_{x + 2} x = \frac{{\log}_{2} x}{{\log}_{2} \left(x + 2\right)}$
$\left({\log}_{2} \frac{2 x - 3}{\log} _ 2 x\right) - {\log}_{2} \frac{x + 6}{{\log}_{2} x} + {\log}_{2} \frac{x + 2}{\log} _ 2 x = 1$
$\frac{{\log}_{2} \left(2 x - 3\right) - {\log}_{2} \left(x + 6\right) + {\log}_{2} \left(x + 2\right)}{{\log}_{2} x} = 1$
${\log}_{2} \left(2 x - 3\right) - \textcolor{red}{{\log}_{2} \left(x + 6\right) + {\log}_{2} \left(x + 2\right)} = {\log}_{2} x$
$\textcolor{g r e e n}{{\log}_{2} \left(2 x - 3\right) - {\log}_{2} \left(x + 6\right) \cdot \left(x + 2\right)} = {\log}_{2} x$
${\log}_{2} \left(\frac{2 x - 3}{\left(x + 6\right) \left(x + 2\right)}\right) = {\log}_{2} x$
$\frac{2 x - 3}{\left(x + 6\right) \left(x + 2\right)} = x$
$x \left({x}^{2} + 2 x + 6 x + 12\right) = 2 x - 3$
${x}^{3} + 2 {x}^{2} + 6 {x}^{2} + 12 x - 2 x + 3 = 0$
${x}^{3} + 8 {x}^{2} + 10 x + 3 = 0$