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Math Help - Help with an exponential/logarithmic equation

  1. #1
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    Help with an exponential/logarithmic equation

    I am trying to solve for x in the equation:

    2(5)^x = 3^(x+1)

    I know the answer should be x=0.79375 because I've graphed them as functions and found the intersect, as well as run the equation through wolfram.

    However, I try to find this by solving for myself and I always seem to come up with x=0.51815.

    I have:
    2(5)^x = 3^(x+1)

    xLog(25) = (x+1)Log3

    xLog25 = xLog3 + Log3

    xLog25 - xLog3 = Log3

    x(Log25 - Log3) = Log3

    x(Log25 - Log3)/(Log25 - Log3) = Log3/(Log25 - Log3)

    x = Log3/(Log25 - Log3) which is 0.51815

    I must be making a mistake somewhere but I just don't see it. Any help is appreciated, thanks!
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  2. #2
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    Re: Help with an exponential/logarithmic equation

    Quote Originally Posted by TypicalNinja View Post
    I have:
    2(5)^x = 3^(x+1)

    xLog(25) = (x+1)Log3
    Where did the 25 come from? Was it \ln 2\times 5^x = x\ln 2\times 5  = x\ln 25 ?
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  3. #3
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    Re: Help with an exponential/logarithmic equation

    Quote Originally Posted by pickslides View Post
    Where did the 25 come from? Was it \ln 2\times 5^x = x\ln 2\times 5  = x\ln 25 ?

    No, the equation given to me was simply 2 x 5^x = 3^(X+1), I put both side into logarithmic form to try and solve it (context is that this is what I'm suppose to do for every question in this assignment).

    I skipped out a step when describing what I did, my first step was to log both sides, so the equation became 2log5^x = log3^(x+1). Then I used the power law to take the exponent of the logs and change it to the coefficient. So the equation became 2xlog5 = (X+1)log3. And I also did the reverse on the left side, took the coefficient 2 and brought it back into the log, giving me xlog5^2, which is xlog25. Am I making a mistake somewhere in that process?
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Help with an exponential/logarithmic equation

    Taking the logs of both sides is a good start, so you get:
    \log(2\cdot 5^x)=\log(3^{x+1})
    \Leftrightarrow \log(2)+x\log(5)=(x+1)\log(3)
    Bring the variable(s) x and the constants to one side.
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  5. #5
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    Re: Help with an exponential/logarithmic equation

    deepashree k s


    yes, u did a mistake in first step.

    u take log for 2(5)^x that becomes log2+xlog5.by applying this u slove the problem and you can get ur result 0.7937.
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  6. #6
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    Re: Help with an exponential/logarithmic equation

    Quote Originally Posted by Siron View Post
    Taking the logs of both sides is a good start, so you get:
    \log(2\cdot 5^x)=\log(3^{x+1})
    \Leftrightarrow \log(2)+x\log(5)=(x+1)\log(3)
    Bring the variable(s) x and the constants to one side.
    Oh I see, my mistake was not applying the log to the entire left side of the equation (I only applied the log to the 5^x, and left the 2 alone during this step). Basic algebra error, I had a feeling I was messing up something basic like that. Thanks very much!

    EDIT: Thanks deepashree as well!
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