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!

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.

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

Re: Fairly basic natural log query (C3)

Quote:

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?

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

$\displaystyle e^{-3x}=27$

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