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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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: Then edit: oops, beaten to it! My Latex skills are too slow.
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, is just the solution to . So let then or equivalently so that
Of course neither nor are fully simplified
Last edited by e^(i*pi); Jan 11th 2010 at 01:18 PM. Reason: that's what I get for being a smart-arse >.<
why would you say that is simplified?? it doesn't look any like any reduced form? just being cranky.
Woah how does ln(1/8) simplify to 3ln(2) and so forth?
Originally Posted by BugzLooney Woah how does ln(1/8) simplify to 3ln(2) and so forth? and
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.
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.
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 , .
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