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Logarithmic Equation

  1. Nov 14, 2008 #1
    Just wondering if someone can tell me if I got this correct. It was a problem on my midterm a month ago and im going over the probs I got wrong, and I dont know the correct answers.
    1. The problem statement, all variables and given/known data

    4(x+1)= 8(x-1)

    2. Relevant equations



    3. The attempt at a solution
    log 4(x+1) = log 8(x-1)

    (x+1) log 4 = (x-1) log 8

    x log 4 + log 4 = x log 8 - log 8 (sub/add to get x on one side)

    log 8 + log 4 = x log 8 - x log 4

    log 8 + log 4 = x(log 8 - log 4) (divide to get x)

    log 8 + log 4 = X
    ---------------
    log 8 - log 4

    x= 5
     
  2. jcsd
  3. Nov 14, 2008 #2

    Dick

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    I guess that works, though you didn't really say how you changed your final log expression into 5. But you can make your life a bit easier by picking a convenient base for the logarithms. In this case I'd pick log_2 (log base 2) and use that. So your second line just turns into 2(x+1)=3(x-1).
     
  4. Nov 14, 2008 #3
    ^
    well, that would make it easier, now wouldnt it? prob why i got it wrong on the exam. i totally forgot about that.

    To get 5, I divided
    (log 8 + log 4)/(log 8 - log 4)
    which is
    1.505149978/.3010299957 = 4.999999998

    thank you though for reminding me about the bases. I couldnt do it the way I did it if it was on the non calculator portion of the exam, so Im going to have to remember that. I have another log question, but its a diff problem, so ill make another post.
     
  5. Nov 14, 2008 #4

    Dick

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    You could also do it this way. (log(8)+log(4))=log(8*4)=log(32)=log(2^5)=5*log(2) and (log(8)-log(4))=log(8/4)=log(2). So you can do it without a calculator, though it pays to get rid of the logs early.
     
  6. Nov 14, 2008 #5
    ah^
    that is like the other question i posted. I forgot on the exam that addition is multiplication. Well, not on the expanding log questions but on solve for x questions.
     
  7. Nov 14, 2008 #6

    symbolipoint

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    Use of logarithms is not necessary for that example.
    (2^2)^(x+1) Left Hand Side.
    (2^3)^(x-1) Right Hand Side.

    Skipping one step,
    2^(2x+2) = 2^(3x-3)
    Left and Right sides having the same base, the exponents are equal.
     
  8. Nov 14, 2008 #7

    Dick

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    There's more than one way to skin a cat.
     
  9. Nov 15, 2008 #8

    CompuChip

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    Yep, here's yet another one;
    4^(x+1) = 8^(x-1)
    16 * 4^(x - 1) = 8^(x - 1)
    16 = (8/4)^(x - 1) = 2^(x - 1)
    x - 1 = 4

    Of course, most methods mentioned are not general, because they rely on the convenient fact that 4= 2^2 and 8 = 2^3.
     
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