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Thread: Logs problem

  1. #1
    Junior Member
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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.

    $\displaystyle 3^{x} \times 4^{2x+1} = 6^{x+2} $ show that $\displaystyle 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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    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.

    $\displaystyle 3^{x} \times 4^{2x+1} = 6^{x+2} $ show that $\displaystyle 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
    $\displaystyle 3^{x} \times 4^{2x+1} = 6^{x+2}$

    Taking logs:

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

    Splitting logs:

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

    Using the power rule on each individual term:

    $\displaystyle xln(3)+(2x+1)ln4=(x+2)ln6$

    Isolating the x's:

    $\displaystyle xln(3)+2xln(4)-xln(6)=2ln(6)-ln(4)$

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

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

    See if you can proceed, using the rules that:

    $\displaystyle log(ab)=log(a)+log(b)$

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

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

    I would start by first rewriting $\displaystyle 4^{2x+1}$ and $\displaystyle 6^{x+2}$ into $\displaystyle 4^{2x} \times 4^1$ and $\displaystyle 6^x \times 6^2$ 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:

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

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

    Rearrange terms and get:

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

    Simplify:

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

    $\displaystyle 8^x = 9$

    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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