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

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Answer 1 · 2015-12-13T12:42:59

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

We will use the following properties:

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

$\Rightarrow ln (e^{- x}) = ln (5^{2 x})$ $=> ln(e^(-x)) = ln(5^(2x))$

$\Rightarrow - x ln (e) = 2 x ln (5)$ $=> -xln(e) = 2xln(5)$

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

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

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

$\Rightarrow x = 0$ $=> x = 0$

How do you solve e−x=52xe^(-x) = 5^(2x) ?