How do you solve $4^{x} = \sqrt{5^{x + 2}}$ $4^x=sqrt(5^(x+2))$ ?

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How do you solve $4^{x} = \sqrt{5^{x + 2}}$ $4^x=sqrt(5^(x+2))$ ?

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Answer 1 · 2016-11-17T13:20:23

$x = \frac{2 log (5)}{4 log (2) - log (5)}$ $x = (2log(5))/(4log(2)-log(5))$

Explanation:

Squaring both sides

$4^{2 x} = 5^{2} 5^{x}$ $4^(2x)=5^2 5^x$ or
$16^{x} = 5^{x} 5^{2}$ $16^x=5^x 5^2$ or
${(\frac{16}{5})}^{x} = 5^{2}$ $(16/5)^x=5^2$ now applying $log$ $log$ to both sidexs
$x (log (16) - log (5)) = 2 log (5)$ $x(log(16)-log(5))=2log(5)$ then
$x = \frac{2 log (5)}{4 log (2) - log (5)}$ $x = (2log(5))/(4log(2)-log(5))$

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How do you solve $4^{x} = \sqrt{5^{x + 2}}$ $4^x=sqrt(5^(x+2))$ ?

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