How do you solve ${log}_{2} (\frac{- 9 x}{2 x^{2} - 1}) = 1$ $log_2((-9x)/(2x^2-1)) = 1$ ?

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How do you solve ${log}_{2} (\frac{- 9 x}{2 x^{2} - 1}) = 1$ $log_2((-9x)/(2x^2-1)) = 1$ ?

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Answer 1 · 2016-05-30T09:04:40

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

Explanation:

${log}_{2} (\frac{- 9 x}{2 x^{2} - 1}) = {log}_{2} (- 9 x) - {log}_{2} (2 x^{2} - 1) = 1 = {log}_{2} 2$ $log_2((-9x)/(2x^2-1)) = log_2(-9x)-log_2(2x^2-1) = 1 = log_2 2$

Since ${log}_{2} (x)$ $log_2(x)$ is a one-one function (as a Real function), we require:

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

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

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

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

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

Use the quadratic formula to find roots:

$x = \frac{- 9 \pm \sqrt{9^{2} - (4 \cdot 4 \cdot - 2)}}{2 \cdot 4}$ $x = (-9+-sqrt(9^2-(4*4*-2)))/(2*4)$

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

$= \frac{- 9 \pm \sqrt{123}}{8}$ $=(-9+-sqrt(123))/8$

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

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

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How do you solve ${log}_{2} (\frac{- 9 x}{2 x^{2} - 1}) = 1$ $log_2((-9x)/(2x^2-1)) = 1$ ?

1 Answer

Explanation:

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How do you solve ${log}_{2} (\frac{- 9 x}{2 x^{2} - 1}) = 1$ $log_2((-9x)/(2x^2-1)) = 1$ ?