How do you solve $3^ { x } = \sqrt { 5^ { x - 2} }$ ?

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Answer 1 · 2017-05-25T21:10:59

$x=log_(5/9)(25)$

Let's rewrite the equation:
$3^x=(5^(x-2))^(1/2)$

$(3^x)^2=(5^(x-2))^(1/2*2)$
$3^(2x)=(5^x)/5^2$
$(5^x)/3^(2x)=25$
$(5/9)^x=25$

$x=log_(5/9)(25)~~-5.48$

If you are unfamiliar with the concept of logarithm, here's a quick overview:

If we have $b^x=y$
Then, $x=log_(b)(y)$ , pronounced "log base b of y."

How do you solve 3^ { x } = \sqrt { 5^ { x - 2} }3^ { x } = \sqrt { 5^ { x - 2} }?