Question

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

Answer 1

0.04

Explanation:

We have,

`e^(-2ln5)= e^(ln(5^-2))=5^-2=1/5^2=1/25=0.04`

Using the following properties of logarithms and exponentials,

`1.` `n*ln` `(m)=ln` `(m^n)` ; `color(blue)(Here)` `color(blue)(Put)` `color(blue)(n=-2)` `color(blue)(and)` `color(blue)(m=5)`

and

`2.` `e^(ln(a))=a` ; `color(blue)(Here)` `color(blue)(Put)` `color(blue)(a=5^-2)`