# Quick Question about Logarithms and Exponents

• May 1st 2013, 10:44 PM
lukasaurus
Quick Question about Logarithms and Exponents
I have an assignment question looks like this

$2e^-^2^x) = 5$

I figured, first time around, that I could just apply natural logarithm to both, and get
$-4x = ln(5)$

and then x = ln(5)/-4

This does not result in the correct answer.

Why? I can get the correct answer (by dividing both sides by two initially)

If I have 2e^(-2x) and apply the natural logarithm to it, what does it become?
• May 1st 2013, 10:52 PM
MarkFL
Re: Quick Question about Logarithms and Exponents
Divide through by 2 first, then convert from exponential to logarithmic form.

If you take the natural log of both sides, you have:

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

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

$-2x=\ln(5)-\ln(2)=\ln\left(\frac{5}{2} \right)$

Same as if you had just divided through by 2 initially.