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Math Help - Exponential and Logarithm Equations

  1. #1
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    Exponential and Logarithm Equations

    Hello everybody,

    I'm having trouble with this problem.

    3*2^x^-^2 = 6-2^x^-^2

    This homework question has me completely stumped, my teacher said that we should first solve for 2^x^-^2 then take the log, but I have no clue what she means, and there is no similar problems in the book. I wish I would've asking her to clarify this. can anyone help me?

    Thank You
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  2. #2
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    \displaystyle 3\times 2^{x-2} = 6-2^{x-2}

    \displaystyle 3\times 2^{x-2} - 6+2^{x-2}= 0

    Make u = 2^{x-2}

    \displaystyle 3\times u - 6+u= 0

    Can you solve this?
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  3. #3
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    Oh! it's the u substitution yes I can do that, sorry for the late response I went to sleep right after I posted. I will try to solve it now, Thank you
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  4. #4
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    Quote Originally Posted by pickslides View Post
    \displaystyle 3\times 2^{x-2} = 6-2^{x-2}

    \displaystyle 3\times 2^{x-2} - 6+2^{x-2}= 0

    Make u = 2^{x-2}

    \displaystyle 3\times u - 6+u= 0

    Can you solve this?
    Actually now that I think of it, we only used u-substitution for integrals thats why I didn't think of that. Are you sure its not another way to do this?

    Thank You
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  5. #5
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    There are many different ways to solve equations but that is surely the simplest. I cannot imagine why you would not want to make an obvious substitution. Typically, one learns to solve equations with substitution, among other methods, several years before integration!
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    There are many different ways to solve equations but that is surely the simplest. I cannot imagine why you would not want to make an obvious substitution. Typically, one learns to solve equations with substitution, among other methods, several years before integration!
    Yes, I figured out how to do it shortly after posting that message.

    \displaystyle 3u +u= 6

    \displaystyle u(3 +1)= 6

    \displaystyle u(4)= 6

    \displaystyle u= 6/4

    Then I plug the 2^x^-^2 back in and use log.

    Quite easy, I just didn't see it until pickslides pointed it out. Sometimes thats how math works LOL

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