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

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Answer 1 · 2015-12-13T12:42:59

Apply properties of logarithms to find that $x=0$

We will use the following properties:

$e^(-x) = 5^(2x)$

$=> ln(e^(-x)) = ln(5^(2x))$

$=> -xln(e) = 2xln(5)$

$=> -x = 2ln(5)x$

$=> 2ln(5)x+x = 0$

$=> x(2ln(5)+1) = 0$

$=> x = 0$

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