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.

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- Jan 11th 2010, 12:56 PMBugzLooneyWhy 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 PMpickslides
$\displaystyle x=-\ln(8) \implies x=\ln(8)^{-1} \implies x=\ln\left(\frac{1}{8}\right)$

All good? - Jan 11th 2010, 01:05 PMpomp
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 PMDrexel28
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 PMe^(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 PMbigwave
why would you say that is simplified??

it doesn't look any like any reduced form?

just being cranky. - Jan 11th 2010, 02:06 PMBugzLooney
Woah how does ln(1/8) simplify to 3ln(2) and so forth?

- Jan 11th 2010, 02:08 PMDrexel28
- Jan 11th 2010, 02:24 PMBugzLooney
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 PMDrexel28
- Jan 12th 2010, 04:45 AMHallsofIvy