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Math Help - logarithm question

  1. #1
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    logarithm question

    I'm bad at maths so how do you do this question?

    (3x)^lg(3) = (4x)^lg(4)

    Find the exact value of 'x'

    Thanks in advance
    Last edited by zResistance; December 5th 2011 at 06:36 AM.
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  2. #2
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    Re: logarithm question

    Quote Originally Posted by zResistance View Post
    I'm bad at maths so how do you do this question?

    (3x)^lg(3) = (4x)^lg(4)

    Find the exact value of 'x'

    Thanks in advance
    \log_3{(3x)^{\ln(3)}}=\log_3{(4x)^{\ln(4)}} \Rightarrow \ln(3)(1+\log_3{x})=\ln(4)(\log_3{4}+\log_3{x})

    Now solve for \log_3{x} . Final result is : x=\frac{1}{12}
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  3. #3
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    Re: logarithm question

    Thanks ALOT , greatly appreaciated for your help
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  4. #4
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    Re: logarithm question

    Hello, zResistance!

    Even if you're good at maths, this problem is a killer!

    Here's a slightly different approach.


    \text{Solve for }x\!:\;\;(3x)^{\ln 3}\:=\:(4x)^{\ln4}

    Take logs: . \ln\left[(3x)^{\ln 3}\right] \:=\:\ln\left[(4x)^{\ln4}\right]

    n . . . . . . . . \ln3 \ln(3x) \:=\:\ln4 \ln(4x)

    a . . . . \ln 3\big[\ln 3 + \ln x\big] \:=\: \ln4\big[\ln4 + \ln x\big]

    . . . . . (\ln 3)^2 + \ln3\ln x \:=\:(\ln4)^2 + \ln4\ln x

    . . . . \ln4\ln x - \ln3\ln x \:=\:-(\ln4)^2 + (\ln3)^2

    x . . . . (\ln4 - \ln3)\ln x \:=\:-\big[(\ln4)^2 - (\ln3)^2\big]

    x . . . . . . . . . . . . \ln x \:=\:-\frac{(\ln4 - \ln3)(\ln4+\ln3)}{\ln4-\ln3}

    x . . . . . . . . . . . . \ln x \:=\:-(\ln 4+\ln3) \:=\:-\ln(3\cdot4)

    x . . . . . . . . . . . . \ln x \:=\:-\ln12 \:=\:\ln(12)^{-1}

    x . . . . . . . . . . . . \ln x \:=\:\ln\left(\frac{1}{12}\right)

    . . . . . . . . . . . . . . . x \:=\:\frac{1}{12}

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