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Math Help - Solve x using logarithms

  1. #1
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    Post Solve x using logarithms

    I have this problem which I can't seem to work out how to solve. I can get the answer, using Wolfram|Alpha, but I want to know the steps of solving it.

    3^(2x) + 3^x = 20

    I know x = log(4)/log(3) = 1.2618595...

    But how do you work it out? Can someone show me the steps?

    The thing that is really confusing me is the plus in the middle, i can't go: log(3^(2x)) + log(3^x) = log(20) because it doesn't work.
    What do you put in the middle: log(3^(2x)) ? log(3^x) = log(20) or am i tackling this completely wrong?

    No matter how I change/re-arrange the equation, I can't seem to get rid of the plus to put logs into it
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  2. #2
    MHF Contributor

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    Quote Originally Posted by DavidM View Post
    I have this problem which I can't seem to work out how to solve. I can get the answer, using Wolfram|Alpha, but I want to know the steps of solving it.

    3^(2x) + 3^x = 20

    I know x = log(4)/log(3) = 1.2618595...

    But how do you work it out? Can someone show me the steps?

    The thing that is really confusing me is the plus in the middle, i can't go: log(3^(2x)) + log(3^x) = log(20) because it doesn't work.
    That's right: log(a+ b) is NOT equal to log(a)+ log(b)

    What do you put in the middle: log(3^(2x)) ? log(3^x) = log(20) or am i tackling this completely wrong?

    No matter how I change/re-arrange the equation, I can't seem to get rid of the plus to put logs into it
    Yes, you are "tackling this completely wrong". First recognize that 3^{2x}= (3^x)^2 so your equation is the same as (3^x)^2+ <br />
(3^x)= 20. If you let y= 3^x your equation becomes y^2+ y= 20. Solve that quadratic equation for y, then use logarithms to solve 3^x= y.
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