# Fairly basic natural log query (C3)

• Nov 3rd 2012, 06:15 AM
DonGorgon
Fairly basic natural log query (C3)
I'm fairly sure this is the correct answer:

e^-3x = 27

-3x = ln27

x = ln27 / -3

However, in the answers section of my textbook, the given value for the answer is -ln3.
Please can someone help explain how the two answers are related and how to obtain such answer from the equation. I am only familiar with the above method.
Thanks!
• Nov 3rd 2012, 06:24 AM
Plato
Re: Fairly basic natural log query (C3)
Quote:

Originally Posted by DonGorgon
I'm fairly sure this is the correct answer:
e^-3x = 27
-3x = ln27
x = ln27 / -3
However, in the answers section of my textbook, the given value for the answer is -ln3.

$\ln(27)=\ln(3^3)=3\ln(3)$
• Nov 3rd 2012, 06:32 AM
DonGorgon
Re: Fairly basic natural log query (C3)
Quote:

Originally Posted by Plato
$\ln(27)=\ln(3^3)=3\ln(3)$

Cheers, thought as much. However is there any different steps to take, or is it just as you mentioned?
• Nov 3rd 2012, 06:48 AM
Plato
Re: Fairly basic natural log query (C3)
Quote:

Originally Posted by DonGorgon
Cheers, thought as much. However is there any different steps to take, or is it just as you mentioned?

You could do
$e^{-3x}=27$
$e^{-x}=3$ (cube root).