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Math Help - Logarithms

  1. #1
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    Logarithms

    Solve for exact value of x for (3x)^lg3 = (4x)^lg 4
    [Ans = 1/12]

    I took lg on both sides and rearranged to get
    lg 3x / lg 4x = lg 4 / lg 3
    lg 3x / lg 4x = 1.262 (evaluate)
    lg 3x = 1.262 lg 4x (Cross-multiply)
    lg 3x - lg 4x^1.262 = 0
    lg (3x/4x^1.262) = 10^0
    3x = 4x^1.262
    3/4 = x^1.262 / x
    Solving, x = 3/4^(1/0.262) = 0.334

    Please advise where I went wrong with my working, coz my answers' not right!
    Thanks !
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  2. #2
    Senior Member Sambit's Avatar
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    I think you have taken the log with base 10. The answer 1/12 is obtained when 'e' is used as the base. Perhaps you have done it correct otherwise. I am giving here another approach of solution:
    Rewrite the given equation as 3^{log(3)} x^{log3} = 4^{log4} x^{log4}
    => \frac{3^{ln3}}{4^{ln4}} = x^{ln 4 - ln 3}
    => \frac{3^{ln3}}{4^{ln4}} = x^{ln \frac{4}{3}}
    => 0.489=x^{0.288}
    => x=e^{\frac{ln 0.489}{0.288}}
    => x=0.0833 =\frac{1}{12}
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  3. #3
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    Hello, Drdj!

    If we are very careful, we can solve it without a calculator.


    \text{Solve for }x\!:\;\;(3x)^{\log3} \:=\: (4x)^{\log4}

    \text{Answer: }\:\frac{1}{12}

    Take logs (base 10): . \log(3x)^{\log 3} \:=\:\log(4x)^{\log4}

    . . . . . . . . . . . . . . \log3\!\cdot\!\log(3x) \:=\:\log4\!\cdot\!\log(4x)

    . . . . . . . . . . \log3\left[\log3 + \log x\right] \:=\:\log4\left[\log4 + \log x\right]

    . - - . . . . . . (\log3)^2 + \log3\!\cdot\!\log x \:=\: (\log4)^2 + \log4\!\cdot\!\log x

    . . = . . . \log4\!\cdot\!\log x - \log3\!\cdot\!\log x \:=\:(\log3)^2 - (\log4)^2

    . . . . . . . . . . (\log4 - \log 3)\log x \:=\:-\left[(\log 4)^2 - (\log 3)^2\right]

    . . . . . . . . . . . . . . . . . . . . \log x \:=\:-\dfrac{(\log 4 - \log 3)(\log 4 + \log 3)}{\log 4 - \log 3}

    . . . . . . . . . . . . . . . . . . . . \log x \:=\:-\left(\log 4+\log3)

    . . . . . . . . . . . . . . . . . . . . \log x \:=\:-\log(4\!\cdot\!3)

    . . . . . . . . . . . . . . . . . . . . \log x \:=\:-\log12

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

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

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

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  4. #4
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    Thanks for the no-calculator method !!
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  5. #5
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    Thanks for the alternative method!
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