Algebra Problem of The Day-10

[BigBeachBanana]

[math]\large 2\log_2x-\log_2\!\left(x-\frac{1}{2}\right)=\log_33[/math]
Solve for [imath]\large x \in \R.[/imath]

 

[red12dog34]

Spoiler

First of all [imath]x>\displaystyle\frac{1}{2}[/imath].
Using the properties of logarithms we have
[math]\log_2\left(\displaystyle\frac{x^2}{x-\frac{1}{2}}\right)=\log_22\Leftrightarrow\frac{x^2}{x-\frac{1}{2}}=2\Leftrightarrow (x-1)^2=0\Leftrightarrow x=1[/math]

 

[JeffM]

Spoiler

[math] 2 \log_2(x) - \log_2 \left ( x - \dfrac{1}{2} \right ) = \log_3(3) = 1 = \log_2(2) \implies \\ 2 \log_2(x) = \log_2 \left ( x - \dfrac{1}{2} \right ) + \log_2(2) = \log_2 \left \{ 2 * \left ( x - \dfrac{1}{2} \right ) \right \} \implies \\ \log_2(x^2) = \log_2(2x - 1) \implies x^2 = 2x - 1 \implies x = 1. [/math]