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Solving Log Equations (Different Bases)

  1. Jan 21, 2010 #1
    1. The problem statement, all variables and given/known data

    Only the equation: 2(3^x) = 7(5^x)



    2. Relevant equations

    N/A

    3. The attempt at a solution

    Okay. I've been fine learning the loglaws, until I hit this question. If they were the same base (or at least near the same, ex: 2 & 8) then I could simply change them into the same base, and proceed. But I can't do that for this question - neither can I just throw "log" it all. (Though that's my solution for now - I'm really not sure how I should proceed.)

    So here's what I've tried:

    2(3^x) = 7(5^x)
    log (2(3^x)) = log (7(5^x))
    x log (2(3)) = x log (7(5))
    x log (6) = x log (35)
    x = x ( log (35) / log (6) )
    x = 1.98x

    Well that doesn't work out too well.. I've tried googling the answer - but all I get is examples with the same base, or a base that I can change to make them the same. So how could I solve this?
     
  2. jcsd
  3. Jan 21, 2010 #2

    rock.freak667

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    Homework Helper

    from here

    log (2(3^x)) = log (7(5^x))

    the next line should be

    lg2+lg3x=lg7+lg5x
     
  4. Jan 21, 2010 #3
    Thanks so much. I had learned to put them together, but the thought of splitting them back up never occurred.
     
  5. Jan 22, 2010 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The crucial point is that [itex]2(3^x)[/itex] is NOT [itex](2(3))^x[/itex] so that [itex]ln(2(3^x)[/itex] is NOT x ln(2(3)).
     
  6. Jan 16, 2012 #5
    Here's an alternate solution:
    2(3^x)=7(5^x)
    2/7=(5^x)/(3^x) Divide both sides by 7 and by 3^x
    2/7=(5/3)^x
    log(2/7)=log((5/3)^x)
    log(2/7)=x log(5/3)
    log(2/7)/log(5/3)=x
     
    Last edited: Jan 16, 2012
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