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Solve algebraically: e^4x-1=24
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\[e^{4x}-1=24\\or~e^{4x-1}=24\\or~(e^4)x-1=24\]
which one?
\[e ^{4x-1}=24 \\ e ^{4x-1}=e ^{24} \\ 4x =1 + e ^{24} \\ x=\frac{ 1+ e ^{24} }{ 4 } \]
the second one
ok, take luis solution. @Luis_Rivera thanks a lot
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nope, he's wrong.
4x -1 =ln24 4x = ln24-1 \[x = \frac{ln 24 -1}{4}\]
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
so take my step, it's your answer, right? and step to get is there. need explanation?
No I got it now, I understand it better :) Thanks, you were a great help
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yw
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