How do you solve ${log}_{2} (x + 2) - {log}_{2} x = 3$ $log_2(x+2) - log_2x = 3$ ?

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How do you solve ${log}_{2} (x + 2) - {log}_{2} x = 3$ $log_2(x+2) - log_2x = 3$ ?

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Answer 1 · 2016-03-31T06:39:16

$x = \frac{2}{7}$ $x = 2/7$

Explanation:

Using log a - log b = $log (\frac{a}{b})$ $log(a/b)$ ,
${log}_{2} (\frac{x + 2}{x}) = 3$ $log_2((x+2)/x) = 3$
Inversely, $(\frac{x + 2}{x}) = 2^{3} = 8$ $((x+2)/x)=2^3=8$
7x = 2.

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How do you solve ${log}_{2} (x + 2) - {log}_{2} x = 3$ $log_2(x+2) - log_2x = 3$ ?

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Explanation:

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How do you solve ${log}_{2} (x + 2) - {log}_{2} x = 3$ $log_2(x+2) - log_2x = 3$ ?