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Thread: Solving with logs

  1. #1
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    Solving with logs

    How do you solve this problem leaving logs in there?

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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Togechu64 View Post
    How do you solve this problem leaving logs in there?

    $\displaystyle \ln{6^{7x+1}}=\ln{7^{6x-1}}$

    $\displaystyle (7x+1)\ln(6)=(6x-1)\ln(7)$

    Now distribute the lns, then solve for x.
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  3. #3
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    I tried that, but I'm not sure where to go after I distribute? :/
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  4. #4
    Super Member Bacterius's Avatar
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    $\displaystyle (7x+1)\ln(6)=(6x-1)\ln(7)$

    Put all logs on a side, and the $\displaystyle x$ terms on the other side (distribute the ln's, that is) :

    $\displaystyle \frac{7x+1}{6x - 1} = \frac{\ln(7)}{\ln(6)}$

    Say $\displaystyle a = \frac{\ln(7)}{\ln(6)}$ (for ease of manipulation and handling) :

    $\displaystyle \frac{7x + 1}{6x - 1} = a$

    Get rid of the fraction by multiplying each side by $\displaystyle (6x - 1)$.

    $\displaystyle 7x + 1 = a(6x - 1)$

    Then expand and solve for $\displaystyle x$ (then substitute for $\displaystyle a$)

    Does it help ?
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  5. #5
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    yes it does, thank you very much!!
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