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Mathematics 22 Online
OpenStudy (anonymous):

Solve algebraically: e^4x-1=24

OpenStudy (loser66):

\[e^{4x}-1=24\\or~e^{4x-1}=24\\or~(e^4)x-1=24\]

OpenStudy (loser66):

which one?

OpenStudy (anonymous):

\[e ^{4x-1}=24 \\ e ^{4x-1}=e ^{24} \\ 4x =1 + e ^{24} \\ x=\frac{ 1+ e ^{24} }{ 4 } \]

OpenStudy (anonymous):

the second one

OpenStudy (loser66):

ok, take luis solution. @Luis_Rivera thanks a lot

OpenStudy (loser66):

nope, he's wrong.

OpenStudy (loser66):

4x -1 =ln24 4x = ln24-1 \[x = \frac{ln 24 -1}{4}\]

OpenStudy (anonymous):

I think I get it now Loser66... I have the answer key for this problem, I just didn't know how to go about getting the answer. The answer key states x=1+In24/4

OpenStudy (loser66):

so take my step, it's your answer, right? and step to get is there. need explanation?

OpenStudy (anonymous):

No I got it now, I understand it better :) Thanks, you were a great help

OpenStudy (loser66):

yw

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