How do you solve ${log}_{2} x + {log}_{2} x + 1 = 2 - {log}_{2} (\frac{1}{3})$ $Log_[2]x + log_[2]x + 1= 2 - log_[2] (1/3)$ ?

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How do you solve ${log}_{2} x + {log}_{2} x + 1 = 2 - {log}_{2} (\frac{1}{3})$ $Log_[2]x + log_[2]x + 1= 2 - log_[2] (1/3)$ ?

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Answer 1 · 2015-12-07T06:24:44

$x = \sqrt{6}$ $x= sqrt 6$

Explanation:

Given ${log}_{2} x + {log}_{2} x + 1 = 2 - {log}_{2} (\frac{1}{3})$ $log_2x + log_2x+ 1= 2-log_2(1/3)$

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

${log}_{2} x + {log}_{2} x + {log}_{2} (\frac{1}{3}) = 2 - 1$ $log_2x + log_2x +log_2(1/3) = 2-1$

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

${log}_{2} x + {log}_{2} x + {log}_{2} (\frac{1}{3}) = 1$ $log_2x + log_2x +log_2(1/3) = 1$
${log}_{2} (x \cdot x \cdot \frac{1}{3}) = 1$ $log_2(x*x*1/3)=1$
${log}_{2} (\frac{1}{3} x^{2}) = 1$ $log_2 (1/3 x^2) = 1$
Change logarithmic equation to exponential form since
${log}_{a} x = y \Leftrightarrow a^{y} = x$ $log_ax = y hArr a^y = x$
$2^{1} = \frac{1}{3} x^{2}$ $2^1 = 1/3 x^2$
$6 = x^{2}$ $6= x^2$
$x = \pm \sqrt{6}$ $x= +-sqrt6$

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

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How do you solve ${log}_{2} x + {log}_{2} x + 1 = 2 - {log}_{2} (\frac{1}{3})$ $Log_[2]x + log_[2]x + 1= 2 - log_[2] (1/3)$ ?

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Explanation:

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How do you solve log2x+log2x+1=2−log2(13)Log_[2]x + log_[2]x + 1= 2 - log_[2] (1/3)?

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How do you solve ${log}_{2} x + {log}_{2} x + 1 = 2 - {log}_{2} (\frac{1}{3})$ $Log_[2]x + log_[2]x + 1= 2 - log_[2] (1/3)$ ?