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Math Help - question on log expression

  1. #1
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    question on log expression

    Dear all experts,

    Would you all please kindly enlighten me with the following equation:

    3(power of)2x+5 = 28(power of)x+1


    after having a few goes at this equation i have become stuck would any be able to show me the best way to approach it?

    Thank you so much for all your help.
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  2. #2
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    e^(i*pi)'s Avatar
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    Re: question on log expression

    Use your exponent laws to split the exponent addition into multiplication: a^{b+c} = a^ba^c

     3^{2x} \cdot 3^5 = 28^x \cdot 28^1


    We also know that a^{bc} = (a^b)^c so we can say that 3^{2x} = (3^2)^x = 9^x


    Collect terms with x on the LHS and those without on the RHS: \dfrac{9^x}{28^x} = \dfrac{28}{3^5}


    We also know that \dfrac{a^x}{b^x} = \left(\dfrac{a}{b}\right)^x and hence the LHS is equal to \left(\dfrac{9}{28}\right)^x

    Overall therefore: \left(\dfrac{9}{28}\right)^x = \dfrac{28}{3^5}

    Can you solve for x using logarithms from here?


    ==========================================

    Edit: An alternative method is to use factorisation.

    Take the log of both sides directly:

    (2x+5)\ln(3) = (x+1)\ln(28)


    Expand the brackets:  2x\ln(3) + 5\ln(3) = x\ln(28) + \ln(28)

    Collect like terms: 2x\ln(3) - x\ln(28) = \ln(28) - 5\ln(3)

    Factor the LHS: x(2\ln(3) - \ln(28)) = \ln(28)-5\ln(3)

    Divide through to solve for x


    Both methods are good so use your favourite choice. They will probably be expressed differently but they're both the same answer 2\ln(3) = \ln(3^2) = \ln(9)
    Last edited by e^(i*pi); August 20th 2011 at 03:49 AM. Reason: see post
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  3. #3
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    Re: question on log expression

    Janice, tattoo this on your wrist: if a^x = b, then x = log(b) / log(a)
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