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Math Help - Solve the following equation

  1. #1
    Junior Member
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    Solve the following equation

    Having trouble getting started on a few of these.

    1. 5^(x-10) = 125^x

    2. (3/4) log x ^(3/4) + 9 = 0

    thanks again!
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  2. #2
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    Hello..

    1) 125^x=5^(3x) => x-10=3x => x=-5

    2) 3/4 * log x ^(3/4) + 9 = 0 => 3/4*3/4 * log x = -9 => 1/16 * log x = -1 => log x = -16
    if log = ln => x=e^(-16)
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  3. #3
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    Hello, Hapa!

    1)\;\;5^{x-10} \:=\: 125^x
    We have: . 5^{x-10} \;=\;\left(5^3\right)^x \quad\Rightarrow\quad 5^{x-10} \:=\:5^{3x}


    Equate exponents: . x-10 \:=\:3x\quad\hdots\;\text{ etc.}




    2)\;\;\tfrac{3}{4}\log\left(x ^{\frac{3}{4}}\right) + 9 \:=\: 0

    We have: . \tfrac{3}{4}\log\left(x^{\frac{3}{4}}\right) \:=\:-9 \quad\Rightarrow\quad \tfrac{3}{4}\cdot\tfrac{3}{4}\log(x) \:=\:-9 \quad\Rightarrow\quad \tfrac{9}{16}\log(x) \:=\:-9


    . . . . . . \log(x) \:=\:-16 \quad\Rightarrow\quad x \:=\:10^{-16}



    Edit: Too slow ... again!
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  4. #4
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    soroban, how do you do to write the equations so beautiful ?
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  5. #5
    Newbie
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    May 2009
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    he used mathtype to write it like that
    and here is another way to solve:

    5^(x-10)= 125^x

    5^(x-10)= 5^(3x)

    Take "log" of both sides:

    (x-10)log5= (3x)log5

    dividing both sides by log5:

    x-10= 3x

    x= -5
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