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

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Answer 1 · 2016-05-30T09:04:40

$x=(-9-sqrt(123))/8$

$log_2((-9x)/(2x^2-1)) = log_2(-9x)-log_2(2x^2-1) = 1 = log_2 2$

Since $log_2(x)$ is a one-one function (as a Real function), we require:

$(-9x)/(2x^2-1) = 2$

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

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

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

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

$x = (-9+-sqrt(9^2-(4*4*-2)))/(2*4)$

$=(-9+-sqrt(81+32))/8$

$=(-9+-sqrt(123))/8$

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

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

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