How do you simplify $e^{- 2 ln 5}$ $e^(-2ln5)$ ?

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How do you simplify $e^{- 2 ln 5}$ $e^(-2ln5)$ ?

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Answer 1 · 2017-05-30T09:00:30

0.04

Explanation:

We have,

$e^{- 2 ln 5} = e^{ln (5^{- 2})} = 5^{- 2} = \frac{1}{5^{2}} = \frac{1}{25} = 0.04$ $e^(-2ln5)= e^(ln(5^-2))=5^-2=1/5^2=1/25=0.04$

Using the following properties of logarithms and exponentials,

$1 .$ $1.$ $n \cdot ln$ $n*ln$ $(m) = ln$ $(m)=ln$ $(m^{n})$ $(m^n)$ ; $H e r e$ $color(blue)(Here)$ $P u t$ $color(blue)(Put)$ $n = - 2$ $color(blue)(n=-2)$ $and$ $color(blue)(and)$ $m = 5$ $color(blue)(m=5)$

and

$2 .$ $2.$ $e^{ln (a)} = a$ $e^(ln(a))=a$ ; $H e r e$ $color(blue)(Here)$ $P u t$ $color(blue)(Put)$ $a = 5^{- 2}$ $color(blue)(a=5^-2)$

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How do you simplify $e^{- 2 ln 5}$ $e^(-2ln5)$ ?

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How do you simplify e−2ln5e^(-2ln5)?

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How do you simplify $e^{- 2 ln 5}$ $e^(-2ln5)$ ?