How do you solve $7 (2^{x}) = 5^{x - 2}$ $7 (2^x) = 5^(x-2)$ ?

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How do you solve $7 (2^{x}) = 5^{x - 2}$ $7 (2^x) = 5^(x-2)$ ?

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Answer 1 · 2016-03-30T04:15:35

x = $\frac{log 175}{log 2.5}$ $log 175/log 2.5$

Explanation:

$5^{x - 2} = 5^{x} 5^{- 2} = \frac{5^{x}}{5^{2}} = \frac{5^{x}}{25}$ $5^(x-2) = 5^x5^(-2)= 5^x/5^2=5^x/25$ .
So, $\frac{5^{x}}{2^{x}} = 175$ $5^x/2^x=175$ .
${(\frac{5}{2})}^{x} . = 175$ $(5/2)^x.= 175$ .
$x log 2.5 = log 175$ $x log 2.5 = log 175$
$x = \frac{log 175}{log 2.5}$ $x=log 175 /log 2.5$

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