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anonymous

  • one year ago

Dylan solved the exponential equation 3x+1 = 15 and his work is shown below. What is the first step he did incorrectly? Step 1: log3x+1 = log15 Step 2: 3 log(x + 1) = log15 Step 3: log(x + 1) = log 15 over 3 Step 4: log(x + 1) = 0.3920304 Step 5: ln[log(x + 1)] = ln0.3920304 Step 6: x + 1 = −0.936415 Step 7: x = −1.936415 I know it's either step 2 or step 3. This is just all so confusing to me...

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  1. anonymous
    • one year ago
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    \(3^{x+1}=15\) ?

  2. anonymous
    • one year ago
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    x+1 is in the exponent, like that?

  3. anonymous
    • one year ago
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    yes

  4. anonymous
    • one year ago
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    \(3^{x+1}=15\) I would divide by 3 first, \(3^{x}=5\) then, log base 3, \(\log_33^{x}=\log_35\) \(x\log_33=\log_35\) \(x=\log_35\)

  5. anonymous
    • one year ago
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    that is what I see as the quickest approach to the exact solution.

  6. anonymous
    • one year ago
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    oh, I will tell you what though....

  7. anonymous
    • one year ago
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    in other words, \(\Large \log(A^B)~~\ne~~ A~\log(B)\) BUT, \(\Large \log(A^B)~~\ne~~ B~\log(A)\)

  8. anonymous
    • one year ago
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    in other words, the exponent in log, goes outside the log.

  9. anonymous
    • one year ago
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    ah okay

  10. anonymous
    • one year ago
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    the \(\underline{\rm exponent}\)!

  11. anonymous
    • one year ago
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    ok, so what step is incorrect ?

  12. anonymous
    • one year ago
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    Step 3?

  13. anonymous
    • one year ago
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    well, if step 3, right side, says `log(15/3)` and not `(log 15)/3`, then step 3 is incorrect as well.

  14. anonymous
    • one year ago
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    but how can something be correct if mistake is already made?

  15. anonymous
    • one year ago
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    well, step 3, if you assume step 2 is given, is right.

  16. anonymous
    • one year ago
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    OH okay I see now

  17. anonymous
    • one year ago
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    :)

  18. anonymous
    • one year ago
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    Thank you!!

  19. Loser66
    • one year ago
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    @fakeee \(\Large \log(A^B)~~\ne~~ A~\log(B)\) Yes!! \(\Large \log(A^B)~~\ne~~ B~\log(A)\) NNNNNNNNNO

  20. Loser66
    • one year ago
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    Because \(\Large \log(A^B)~~=~~ B~\log(A)\)

  21. Loser66
    • one year ago
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    For example: \(\Large \log(10^2)~~=~~ 2~\log(10)=2\) or \(\Large \log(100)~~=2\)

  22. anonymous
    • one year ago
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    Yeah I meant to say that 2nd eq. is = (not \(\ne\)) apologize... Thanks Loser...(don't mean to offend you xD)

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spraguer (Moderator)
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