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

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Answer 1 · 2016-04-10T20:59:11

I found: $x = \frac{2 ln (5)}{(ln (5) - ln (2))} = 3.513$ $x=(2ln(5))/((ln(5)-ln(2)))=3.513$

I would take the natural log of both sides:

$ln 2^{x} = ln 5^{x - 2}$ $color(red)(ln)2^x=color(red)(ln)5^(x-2)$

then use the fact that ${log x}^{m} = m log x$ $logx^m=mlogx$ and write:
$x ln (2) = (x - 2) ln (5)$ $xln(2)=(x-2)ln(5)$

rearrange:

$x ln (2) - x ln (5) = - 2 ln (5)$ $xln(2)-xln(5)=-2ln(5)$
$x [ln (2) - ln (5)] = - 2 ln (5)$ $x[ln(2)-ln(5)]=-2ln(5)$

and:

$x = \frac{- 2 ln (5)}{(ln (2) - ln (5))} = \frac{2 ln (5)}{(ln (5) - ln (2))} = 3.513$ $x=(-2ln(5))/((ln(2)-ln(5)))=(2ln(5))/((ln(5)-ln(2)))=3.513$

How do you solve 2^x = 5^(x - 2)2x=5x−22^x = 5^(x - 2)?