# Thread: Fairly basic natural log query (C3)

1. ## 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!

2. ## Re: Fairly basic natural log query (C3)

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.
$\displaystyle \ln(27)=\ln(3^3)=3\ln(3)$

3. ## Re: Fairly basic natural log query (C3)

Originally Posted by Plato
$\displaystyle \ln(27)=\ln(3^3)=3\ln(3)$
Cheers, thought as much. However is there any different steps to take, or is it just as you mentioned?

4. ## Re: Fairly basic natural log query (C3)

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
$\displaystyle e^{-3x}=27$
$\displaystyle e^{-x}=3$ (cube root).