# How do you solve Log_[2]x + log_[2]x + 1= 2 - log_[2] (1/3)?

Dec 7, 2015

$x = \sqrt{6}$

#### Explanation:

Given ${\log}_{2} x + {\log}_{2} x + 1 = 2 - {\log}_{2} \left(\frac{1}{3}\right)$

Step 1: Manipulate the equation to have all logarithm on one side, like so

${\log}_{2} x + {\log}_{2} x + {\log}_{2} \left(\frac{1}{3}\right) = 2 - 1$

Step 2: Use the sum to product logarithmic properties
$\log A B = \log A + \log B$

${\log}_{2} x + {\log}_{2} x + {\log}_{2} \left(\frac{1}{3}\right) = 1$
${\log}_{2} \left(x \cdot x \cdot \frac{1}{3}\right) = 1$
${\log}_{2} \left(\frac{1}{3} {x}^{2}\right) = 1$
Change logarithmic equation to exponential form since
${\log}_{a} x = y \Leftrightarrow {a}^{y} = x$
${2}^{1} = \frac{1}{3} {x}^{2}$
$6 = {x}^{2}$
$x = \pm \sqrt{6}$

Only positive answer would work because the domain $\log x$ exist if $x > 0$