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

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Answer 1 · 2016-03-30T04:15:35

x = $log 175/log 2.5$

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

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