# How do you simplify e^(-2ln5)?

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7
Apr 17, 2017

$\frac{1}{25}$

#### Explanation:

${e}^{\textcolor{b l u e}{- 2 \ln 5}}$

When you have $\text{ "color(blue)(-2ln5)" }$ in the exponent,

that is the equivalent of $\text{ "color(blue)(ln5^-2)}$

${e}^{\textcolor{b l u e}{\ln {5}^{-} 2}} \text{ and } {e}^{\ln}$ cancels out to be 1.

So that means you only have ${5}^{-} 2$

Negative exponents mean the answer is $\frac{1}{{5}^{2}}$ = $\frac{1}{25}$

The rules we used when simplifying were: (in order)

$x \ln y = \ln {y}^{x}$

${e}^{\ln} = 1$

${x}^{-} y = \frac{1}{x} ^ y$

Keep in mind these rules for future problems. Good luck!

Then teach the underlying concepts
Don't copy without citing sources
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#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

3
Feb 4, 2018

0.04

#### Explanation:

We have,

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

Using the following properties of logarithms and exponentials,

$1.$ $n \cdot \ln$ $\left(m\right) = \ln$ $\left({m}^{n}\right)$ ; $\textcolor{b l u e}{H e r e}$ $\textcolor{b l u e}{P u t}$ $\textcolor{b l u e}{n = - 2}$ $\textcolor{b l u e}{\mathmr{and}}$ $\textcolor{b l u e}{m = 5}$

and

$2.$ ${e}^{\ln \left(a\right)} = a$ ; $\textcolor{b l u e}{H e r e}$ $\textcolor{b l u e}{P u t}$ $\textcolor{b l u e}{a = {5}^{-} 2}$

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