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

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7
Apr 17, 2017

Answer:

#1/25#

Explanation:

#e^(color(blue)(-2ln5))#

When you have #" "color(blue)(-2ln5)" "# in the exponent,

that is the equivalent of #" "color(blue)(ln5^-2)""#

#e^(color(blue)(ln5^-2))" and "e^ln# cancels out to be 1.

So that means you only have #5^-2#

Negative exponents mean the answer is #1/(5^2)# = #1/25#

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

#xlny = lny^x#

#e^ln = 1#

#x^-y = 1/x^y#

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

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3
Feb 4, 2018

Answer:

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)#

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