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Help me Please. e^x-1-5=5.

Algebra
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Helpl please
Solve for \(e^{x-1}\) and take the natural logarithm of both sides.
I started this problem but its giving me a wrong answer can u please explain this to me?

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Other answers:

Well, tell me, what do you get when you solve for \(e^{x-1}\)?
can anyone help me here please?
\[e^x-1-5=5.\]\[e^x-6=5\]\[e^x=11\]\[ln(e^x)=ln(11)\]\[xln(e)=ln(11)\]\[x=ln(11)\]
Sorry but the answer is wrong answer suppose to be 3.3026
@Jaweria Then you wrote it down wrong.
ok let me fix it
\[e ^{x-1}-5=5\]
\[e^{x-1}-5=5\]Isolate e term.\[e^{x-1}=10\]Natural log of both sides.\[lne^{x-1}=ln(10)\]Properties of logatrithms.\[(x-1)lne=ln(10)\]lne = 1.\[x-1=ln(10)\]Solve for x.\[x=ln(10)+1\]
Sorry but I have one more question if you dont mind ?
\[9^{n+10}+3=81\]
\[9^{n+10}+3=81\]\[9^{n+10}=81-3\]\[ln9^{n+10}=ln78\]\[n+10=ln78/ln9\]\[n=ln78/ln9-10\]
Thanks :)
@Jaweria Properties of Logarithms http://www.andrews.edu/~calkins/math/webtexts/numb17.htm
Thanks for this :)
I m having trouble with this question. \[-2 \log_{5}7x=2 \]
\[-2log_57x=2\]\[log_57x=-1\]\[log_ax=y\]\[a^y=x\]\[5^{-1}=7x\]\[\frac{1}{5}=7x\]\[x=1/35\]
Thanks
I m working on this problem so far I got this but my answer is not matching with my professor's. |dw:1361342182376:dw|
Can anyone please help me ??
Is that \[5*6^{3m} = 20\]?
yes
I'll assume it is, if it isn't, maybe the example will help you. \[5*6^{3m} = 20\]Divide both sides by 5\[6^{3m}=4\]Take natural log of both sides\[3m \ln 6 = \ln 4\]Divide both sides by \( 3 \ln 6\)\[m = \frac{\ln 4}{3 \ln 6}\]
thanks

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