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Math Help - help solving using logarithms..

  1. #1
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    help solving using logarithms..

    i need to solve this equation for x using base 10 logarithms...

    13^(x-3) = 18^(3x)
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Manizzle View Post
    i need to solve this equation for x using base 10 logarithms...

    13^(x-3) = 18^(3x)
    so take the log of both sides

    \log 13^{x - 3} = \log 18^{3x}

    now recall the rule: \log_a (x^n) = n \log_a x

    now proceed
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  3. #3
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    yes i know that rule but i don't know how then to solve for x
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Manizzle View Post
    yes i know that rule but i don't know how then to solve for x
    ok, so that means you got to the point (x - 3) \log 13 = 3x \log 18

    now what is the problem? \log 13 and \log 18 are constants. how would you solve something like (x - 3) \cdot 5 = 3x \cdot 8 ?

    just treat the logs like constants. they are just numbers. solve for x as you would in any regular algebraic equation
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  5. #5
    A riddle wrapped in an enigma
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    Quote Originally Posted by Manizzle View Post
    i need to solve this equation for x using base 10 logarithms...


    13^(x-3) = 18^(3x)
    Sorry, Jhevon. But I had already done the grunt work on this so here's the rest of it.


    13^{x-3}=18^{3x}

    \log(13^{x-3})=\log (18^{3x})

    (x-3)\log 13=(3x)\log 18

    (x-3) \cdot 2.564949357=3x \cdot 2.890371758

    2.564949357x-7.694848071=8.671115274x

    -7.694848071=6.106165917x

    x=\frac{-7.694848071}{6.106165917}

    \boxed{x \approx -1.260176709}
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by masters View Post
    Sorry, Jhevon. But I had already done the grunt work on this so here's the rest of it.


    13^{x-3}=18^{3x}

    \log(13^{x-3})=\log (18^{3x})

    (x-3)\log 13=(3x)\log 18

    (x-3) \cdot 2.564949357=3x \cdot 2.890371758

    2.564949357x-7.694848071=8.671115274x

    -7.694848071=6.106165917x

    x=\frac{-7.694848071}{6.106165917}

    \boxed{x \approx -1.260176709}
    that's fine, Masters. I would have preferred if you left it as log(13) and log(18) though as opposed to finding the decimal approximations.
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