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Calculus: Logarithms

  1. Feb 9, 2005 #1
    Can't get this question, I get the wrong answer:

    4.6*1.06^(2x+3)=5*3^x

    So find x
     
  2. jcsd
  3. Feb 9, 2005 #2

    Galileo

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    Seeing the title of the thread, it seems obvious that you've already started by taking the logarithm on both sides. Show us your work so we know where the problem lies.
     
  4. Feb 9, 2005 #3

    dextercioby

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    HINT:Divide by 4.6 and then apply natural logarithm over both sides of the equation.

    Daniel.
     
  5. Feb 9, 2005 #4
    I did the following:

    ( 4.6*1.06^(2x+3)=5*3^x )/4.6*3^x

    and then cancelled. Dexter then I applied Logarithm and I just want to see how you guys did it.....I got an answer just wrong.....
     
  6. Feb 9, 2005 #5

    dextercioby

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    [tex] 1.06^{2x+3}=\frac{5}{4.6}\cdot 3^{x} [/tex]

    Taking natural logarithm
    [tex] (2x+3)\ln 1.06=\ln\frac{5}{4.6} +x\ln 3 [/tex]

    Solve for "x"...The final answer ain't pretty,by any means.

    Daniel.
     
  7. Feb 9, 2005 #6
    God damnit Daniel.....I'm so retarded.....

    I forgot to Log the right side!!! Such stupid and simple mistakes.....sorry for wasting your time lol.......P.S. I like that special writing you use
     
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