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Logarithms solve for x help

  1. Dec 12, 2008 #1
    1. The problem statement, all variables and given/known data
    Solve for x. State restrictions, if necessary.

    3^(2x) - 3^(x) -12 = 0


    2. Relevant equations



    3. The attempt at a solution

    2xlog3 - log3 = log 12
    x=log 12/log 3

    Doesn't work. I have no idea how to do this.. we didn't learn it. The 2x is throwing me off.
     
  2. jcsd
  3. Dec 12, 2008 #2

    Avodyne

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    Science Advisor

    Re: Logarithms

    Let y=3^x, and try to write the equation in terms of y. Then, solve for y. Then, use y=3^x to solve for x.
     
  4. Dec 12, 2008 #3
    Re: Logarithms

    Let y = 3^(x)

    2y-y-12=0
    y=12
    3^(x)=12
    xlog3=12
    x = 12/log3

    Doesn't work either.
     
  5. Dec 12, 2008 #4

    Avodyne

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    Science Advisor

    Re: Logarithms

    If y=e^x, what is e^(2x) in terms of y? Hint: the answer is not 2y.
     
  6. Dec 12, 2008 #5
    Re: Logarithms

    Let [tex]y=3^x[/tex]
    [tex]
    y^2-y-12=0
    [/tex]

    [tex]
    (y-4)(y+3)=0
    [/tex]

    [tex]
    y=4\ or\ y=-3
    [/tex]

    [tex]
    3^x=4\ or\ 3^x=-3
    [/tex]

    [tex]
    \log 3^x = \log 4
    [/tex]

    [tex]
    x= \frac{\log 4}{\log3}
    [/tex]

    [tex]
    x \approx 1.2618595071429
    [/tex]

    For any value of x, [tex] 3^x\neq-3[/tex], this solution is extraneous (rejected).
     
  7. Dec 12, 2008 #6
    Re: Logarithms

    Are there restrictions btw
     
  8. Dec 12, 2008 #7
    Re: Logarithms

    Not sure what you mean. The only restriction I'm guessing would be that any negative roots you get will never be a solution to an exponential expression, where the exponent is not 1.
     
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