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

May 30, 2016

$x = \frac{- 9 - \sqrt{123}}{8}$

#### Explanation:

${\log}_{2} \left(\frac{- 9 x}{2 {x}^{2} - 1}\right) = {\log}_{2} \left(- 9 x\right) - {\log}_{2} \left(2 {x}^{2} - 1\right) = 1 = {\log}_{2} 2$

Since ${\log}_{2} \left(x\right)$ is a one-one function (as a Real function), we require:

$\frac{- 9 x}{2 {x}^{2} - 1} = 2$

Multiplying both sides by $\left(2 {x}^{2} - 1\right)$ we get:

$- 9 x = 4 {x}^{2} - 2$

Add $9 x$ to both sides and transpose to get:

$4 {x}^{2} + 9 x - 2 = 0$

Use the quadratic formula to find roots:

$x = \frac{- 9 \pm \sqrt{{9}^{2} - \left(4 \cdot 4 \cdot - 2\right)}}{2 \cdot 4}$

$= \frac{- 9 \pm \sqrt{81 + 32}}{8}$

$= \frac{- 9 \pm \sqrt{123}}{8}$

We can discard $\frac{- 9 + \sqrt{123}}{8} > 0$, since it results in $- 9 x < 0$, so the Real logarithm is not defined.

That leaves $x = \frac{- 9 - \sqrt{123}}{8}$ as the only solution of the original equation.