# Why is -ln8 equal ln1/8?

• Jan 11th 2010, 12:56 PM
BugzLooney
Why is -ln8 equal ln1/8?
I solved this problem and couldn't get the last step but apparently it went like this. First e^-x=8
Then ln(8)=-x
So x=-ln(8)

Apparently the answer is then x=ln(1/8)

I just don't know why the negative ln8 is equal to 1/8.
• Jan 11th 2010, 01:02 PM
pickslides
$\displaystyle x=-\ln(8) \implies x=\ln(8)^{-1} \implies x=\ln\left(\frac{1}{8}\right)$

All good?
• Jan 11th 2010, 01:05 PM
pomp
Quote:

Originally Posted by BugzLooney
I solved this problem and couldn't get the last step but apparently it went like this. First e^-x=8
Then ln(8)=-x
So x=-ln(8)

Apparently the answer is then x=ln(1/8)

I just don't know why the negative ln8 is equal to 1/8.

That last line is not true. You need to use the following facts:

$\displaystyle \ln{1} = 0$

$\displaystyle \ln{a} - \ln{b} = \ln{\left(\frac{a}{b}\right)}$

Then

$\displaystyle - \ln{8} = 0 - \ln{8} = \ln{1} - \ln{8} = \ln{\left(\frac{1}{8}\right)}$

edit: oops, beaten to it! My Latex skills are too slow.
• Jan 11th 2010, 01:05 PM
Drexel28
Quote:

Originally Posted by BugzLooney
I solved this problem and couldn't get the last step but apparently it went like this. First e^-x=8
Then ln(8)=-x
So x=-ln(8)

Apparently the answer is then x=ln(1/8)

I just don't know why the negative ln8 is equal to 1/8.

The technical reason...is well technical. But think about it like this, $\displaystyle \ln(x)$ is just the solution to $\displaystyle e^y=x$. So let $\displaystyle \alpha=\ln\left(\frac{1}{8}\right)$ then $\displaystyle e^{\alpha}=\frac{1}{8}$ or equivalently $\displaystyle 8=\frac{1}{e^{\alpha}}=e^{-\alpha}$ so that $\displaystyle \ln(8)=-\alpha\implies -\ln(8)=\alpha=\ln\left(\tfrac{1}{8}\right)$
• Jan 11th 2010, 01:16 PM
e^(i*pi)
Of course neither $\displaystyle -ln(8)$ nor $\displaystyle ln\left(\frac{1}{8}\right)$ are fully simplified (Wink)

$\displaystyle -3ln(2) = 3ln\left(\frac{1}{2}\right)$ (Giggle)
• Jan 11th 2010, 01:23 PM
bigwave
why would you say that is simplified??
it doesn't look any like any reduced form?

just being cranky.
• Jan 11th 2010, 02:06 PM
BugzLooney
Woah how does ln(1/8) simplify to 3ln(2) and so forth?
• Jan 11th 2010, 02:08 PM
Drexel28
Quote:

Originally Posted by BugzLooney
Woah how does ln(1/8) simplify to 3ln(2) and so forth?

$\displaystyle \ln\left(ab\right)=\ln(a)+\ln(b)$ and $\displaystyle \ln\left(a^n\right)=n\ln(a)$
• Jan 11th 2010, 02:24 PM
BugzLooney
I understand everything except for the last post as it relates to mine.

ln(1/8) simplifies to 3ln(2) by that rule? I just don't see it. Though the explanations have been great thus far.
• Jan 11th 2010, 02:33 PM
Drexel28
Quote:

Originally Posted by BugzLooney
I understand everything except for the last post as it relates to mine.

ln(1/8) simplifies to 3ln(2) by that rule? I just don't see it. Though the explanations have been great thus far.

$\displaystyle \ln(8)=\ln\left(2^3\right)=3\ln(2)$
• Jan 12th 2010, 04:45 AM
HallsofIvy
Quote:

Originally Posted by BugzLooney
I solved this problem and couldn't get the last step but apparently it went like this. First e^-x=8
Then ln(8)=-x
So x=-ln(8)

Apparently the answer is then x=ln(1/8)

I just don't know why the negative ln8 is equal to 1/8.

Since $\displaystyle e^{-x}= 8$, $\displaystyle e^x= \frac{1}{8}$.