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Solving logarithmic equations

  1. Apr 13, 2014 #1
    1. The problem statement, all variables and given/known data

    Determine at which points the graphs of the given pair of functions intersect:

    f(x) = 3x and g(x) = 2x2

    2. Relevant equations



    3. The attempt at a solution

    I know I have to equate and solve for x so I converted them to logarithms

    log3x = log2x2

    Don't know if that's right, but I am stuck here, do I use the change of base formula ?
     
  2. jcsd
  3. Apr 13, 2014 #2

    Dick

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    With that subscript in there not clear what you mean by "converted them to logarithms". The correct thing to do is to try to solve the equation log(f(x))=log(g(x)). ##\log 3^x=\log 2^{x^2}##. The 'log' can be any base you like. Just use the rules of logarithms to solve that equation.
     
  4. Apr 13, 2014 #3

    mfb

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    You cannot just replace exponentials by logarithms, it won't work. There is a way to solve it, but then your steps have to be valid transformations.
    That is a good idea, you can do it with the exponentials as well.
     
  5. Apr 13, 2014 #4
    log3x = log2x2

    log3x = 2log2x

    Using the change of base formula

    log2x/log23 = 2log2x............stuck here
     
  6. Apr 13, 2014 #5

    mfb

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    2x2 means 2(x2), not (2x)2, your first step does not work.

    What is log(3x) simplified?
     
  7. Apr 14, 2014 #6
    xlog3 = x^2log2

    log3 = xlog2

    x = log3/log2

    Is that correct ?, also why is the base 10 ?, I thought it was 3 and 2 respectively
     
  8. Apr 14, 2014 #7

    Dick

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    That's part of it. The base doesn't have to be 10. If you take ratio log(3)/log(2) in any base you'll get the same number. Can you say why? More importantly, there is another solution. What is it?
     
  9. Apr 14, 2014 #8
    I think the other solution should be x = 0 as well ?, I'm not too sure about why you get the same number, a bit confused, can you explain that please ?
     
  10. Apr 14, 2014 #9

    HallsofIvy

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    Starting from [itex]3^x= 2^{x^2}[/itex], you can take the logarithm to any base, "10", "e", whatever, and get [itex]log(3^x)= x log(3)= log(2^{x^2})= x^2 log(2)[/itex]. If x is not 0, you can divide both sides by x log(2).
     
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