# Thread: Why is -ln8 equal ln1/8?

1. ## 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.

2. $\displaystyle x=-\ln(8) \implies x=\ln(8)^{-1} \implies x=\ln\left(\frac{1}{8}\right)$

All good?

3. 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.

4. 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 $\displaystyle \ln(8)=-\alpha\implies -\ln(8)=\alpha=\ln\left(\tfrac{1}{8}\right)$

5. Of course neither $\displaystyle -ln(8)$ nor $\displaystyle ln\left(\frac{1}{8}\right)$ are fully simplified

$\displaystyle -3ln(2) = 3ln\left(\frac{1}{2}\right)$

6. why would you say that is simplified??
it doesn't look any like any reduced form?

just being cranky.

7. Woah how does ln(1/8) simplify to 3ln(2) and so forth?

8. 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 $\displaystyle \ln\left(a^n\right)=n\ln(a)$

9. 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.

10. 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.
$\displaystyle \ln(8)=\ln\left(2^3\right)=3\ln(2)$

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

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