How do you solve ${log}_{2} (5 x + 7) - {log}_{2} x = 2$ $log_(2) (5x + 7) - log_(2) x = 2$ ?

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How do you solve ${log}_{2} (5 x + 7) - {log}_{2} x = 2$ $log_(2) (5x + 7) - log_(2) x = 2$ ?

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Answer 1 · 2015-11-02T13:15:57

$x = - 7$ $x = -7$

Explanation:

Note that

${log}_{2} (5 x + 7) - {log}_{2} x = 2$ $log_2 (5x+7)- log_2 x =2$
$\Rightarrow {log}_{2} (5 x + 7) + {log}_{2} (\frac{1}{x}) = 2$ $implieslog_2 (5x+7)+ log_2 (1/x)=2$
$\Rightarrow {log}_{2} (\frac{5 x + 7}{x}) = 2$ $implies log_2((5x+7)/x)=2$
$\Rightarrow \frac{5 x + 7}{x} = 2^{2}$ $implies (5x+7)/(x)= 2^2$
$\Rightarrow 5 x + 7 = 4 x$ $implies 5x+7 =4x$
$\Rightarrow x = - 7$ $implies x = -7$

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How do you solve ${log}_{2} (5 x + 7) - {log}_{2} x = 2$ $log_(2) (5x + 7) - log_(2) x = 2$ ?

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How do you solve ${log}_{2} (5 x + 7) - {log}_{2} x = 2$ $log_(2) (5x + 7) - log_(2) x = 2$ ?