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Math Help - 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?

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

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

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

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

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

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

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

    7x + 1 = a(6x - 1)

    Then expand and solve for x (then substitute for 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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