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Math Help - Logs problem

  1. #1
    Junior Member
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    Oct 2011
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    Logs problem

    Hi there,

    I'm stuck on this logs question and can't find an example in any books that fits this form.

     3^{x} \times 4^{2x+1} = 6^{x+2} show that  x = \frac{\ln 9}{\ln 8}

    I try to break the logs down to different forms but never get anything suitable to allow for a substitution.

    Any pointers?

    Thanks, F
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  2. #2
    Super Member Quacky's Avatar
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    Windsor, South-East England
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    Re: Logs problem

    Quote Originally Posted by FelixHelix View Post
    Hi there,

    I'm stuck on this logs question and can't find an example in any books that fits this form.

     3^{x} \times 4^{2x+1} = 6^{x+2} show that  x = \frac{\ln 9}{\ln 8}

    I try to break the logs down to different forms but never get anything suitable to allow for a substitution.

    Any pointers?

    Thanks, F
    3^{x} \times 4^{2x+1} = 6^{x+2}

    Taking logs:

    ln(3^{x}4^{2x+1})=ln(6^{x+2})

    Splitting logs:

    ln(3^x)+ln(4^{2x+1})=ln(6^{x+2})

    Using the power rule on each individual term:

    xln(3)+(2x+1)ln4=(x+2)ln6

    Isolating the x's:

    xln(3)+2xln(4)-xln(6)=2ln(6)-ln(4)

    Taking out a factor of x and rewriting things using the power rule:

    x(ln(3)+2ln(4)-ln(6))=ln(36)-ln(4)

    See if you can proceed, using the rules that:

    log(ab)=log(a)+log(b)

    log(\frac{a}{b})=log(a)-log(b) and

    log(x^n)=nlog(x)
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  3. #3
    Junior Member Youkla's Avatar
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    Re: Logs problem

    I would start by first rewriting 4^{2x+1} and 6^{x+2} into 4^{2x} \times 4^1 and 6^x \times 6^2 respectively. Then go from there. See if that is enough to get you going.
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  4. #4
    Junior Member Youkla's Avatar
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    Re: Logs problem

    Doing it by using logs like Quacky suggested is one way. What I suggested leads to:

    3^x \times 4^{2x+1} = 6^{x+2}

    3^x \times 4^{2x} \times 4^1 = 6^x \times 6^2

    Rearrange terms and get:

    \frac{3^x \times (4^2)^x}{6^x} = \frac{6^2}{4}

    Simplify:

    (\frac{3 \times 16^}{6})^x = 9

    8^x = 9

    From here, take the natural log of both sides and "Wallah!"
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  5. #5
    Junior Member
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    Re: Logs problem

    Thanks everyone - I'd gone down that avenue of rewriting but must have made a mistake (several times!)

    Thanks again.

    F
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