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Jaweria Group Title

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

  • one year ago
  • one year ago

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  1. Jaweria Group Title
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    Helpl please

    • one year ago
  2. yakeyglee Group Title
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    Solve for \(e^{x-1}\) and take the natural logarithm of both sides.

    • one year ago
  3. Jaweria Group Title
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    I started this problem but its giving me a wrong answer can u please explain this to me?

    • one year ago
  4. yakeyglee Group Title
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    Well, tell me, what do you get when you solve for \(e^{x-1}\)?

    • one year ago
  5. Jaweria Group Title
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    can anyone help me here please?

    • one year ago
  6. stamp Group Title
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    \[e^x-1-5=5.\]\[e^x-6=5\]\[e^x=11\]\[ln(e^x)=ln(11)\]\[xln(e)=ln(11)\]\[x=ln(11)\]

    • one year ago
  7. Jaweria Group Title
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    Sorry but the answer is wrong answer suppose to be 3.3026

    • one year ago
  8. stamp Group Title
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    @Jaweria Then you wrote it down wrong.

    • one year ago
  9. Jaweria Group Title
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    ok let me fix it

    • one year ago
  10. Jaweria Group Title
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    \[e ^{x-1}-5=5\]

    • one year ago
  11. stamp Group Title
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    \[e^{x-1}-5=5\]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.\[x=ln(10)+1\]

    • one year ago
  12. Jaweria Group Title
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    Sorry but I have one more question if you dont mind ?

    • one year ago
  13. Jaweria Group Title
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    \[9^{n+10}+3=81\]

    • one year ago
  14. stamp Group Title
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    \[9^{n+10}+3=81\]\[9^{n+10}=81-3\]\[ln9^{n+10}=ln78\]\[n+10=ln78/ln9\]\[n=ln78/ln9-10\]

    • one year ago
  15. Jaweria Group Title
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    Thanks :)

    • one year ago
  16. stamp Group Title
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    @Jaweria Properties of Logarithms http://www.andrews.edu/~calkins/math/webtexts/numb17.htm

    • one year ago
  17. Jaweria Group Title
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    Thanks for this :)

    • one year ago
  18. Jaweria Group Title
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    I m having trouble with this question. \[-2 \log_{5}7x=2 \]

    • one year ago
  19. stamp Group Title
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    \[-2log_57x=2\]\[log_57x=-1\]\[log_ax=y\]\[a^y=x\]\[5^{-1}=7x\]\[\frac{1}{5}=7x\]\[x=1/35\]

    • one year ago
  20. Jaweria Group Title
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    Thanks

    • one year ago
  21. Jaweria Group Title
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    I m working on this problem so far I got this but my answer is not matching with my professor's. |dw:1361342182376:dw|

    • one year ago
  22. Jaweria Group Title
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    Can anyone please help me ??

    • one year ago
  23. whpalmer4 Group Title
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    Is that \[5*6^{3m} = 20\]?

    • one year ago
  24. Jaweria Group Title
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    yes

    • one year ago
  25. whpalmer4 Group Title
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    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 \( 3 \ln 6\)\[m = \frac{\ln 4}{3 \ln 6}\]

    • one year ago
  26. Jaweria Group Title
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    thanks

    • one year ago
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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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