How do you solve $2^{x} = 5^{x + 6}$ $2^x = 5^(x+6)$ ?

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How do you solve $2^{x} = 5^{x + 6}$ $2^x = 5^(x+6)$ ?

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Answer 1 · 2015-12-02T09:12:33

$6 {log}_{\frac{2}{5}} 5$ $6log_(2/5)5$

Explanation:

$2^{x} = 5^{x + 6}$ $2^x=5^(x+6)$
$\Rightarrow {log}_{5} 2^{x} = x + 6$ $=>log_5 2^x=x+6$ (logarithm definition)
$x {log}_{5} 2 = x + 6$ $xlog_5 2=x+6$ (the logarithm of the power of a number)
Divide both sides into $x$ $x$ :
$\Rightarrow \frac{1}{x} \cdot (x {log}_{5} 2) = \frac{1}{x} \cdot (x + 6)$ $=>1/x*(xlog_5 2)=1/x*(x+6)$
$\Rightarrow \frac{x + 6}{x} = {log}_{5} 2$ $=>(x+6)/x=log_5 2$
Add $- 1$ $-1$ to both sides:
$\frac{x + 6}{x} - 1 = {log}_{5} 2 - 1$ $(x+6)/x-1=log_5 2-1$
$\Rightarrow \frac{6}{x} = {log}_{5} 2 - 1$ $=>6/x=log_5 2-1$
$\Rightarrow x = \frac{6}{{log}_{5} 2 - 1}$ $=>x=6/(log_5 2-1)$
$= \frac{6}{{log}_{5} 2 - {log}_{5} 5}$ $=6/(log_5 2-log_5 5)$
$= \frac{6}{{log}_{5} (\frac{2}{5})}$ $=6/log_5 (2/5)$
$= 6 {log}_{\frac{2}{5}} 5$ $=6log_(2/5)5$

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