Fairly basic natural log query (C3)

Aug 2012
22
0
London
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!
 

Plato

MHF Helper
Aug 2006
22,508
8,664
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.
\(\displaystyle \ln(27)=\ln(3^3)=3\ln(3)\)
 

Plato

MHF Helper
Aug 2006
22,508
8,664
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).