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Math Help - Solving x from log equation

  1. #1
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    Solving x from log equation

    Solving for x

    log(x^2+9)-2logx=log5

    I did:

    10^5 = (x^2 + 9) / x^2

    x's cancel out so I don't have any x to solve for. So clearly I messed up somewhere
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by Devi09 View Post
    Solving for x

    log(x^2+9)-2logx=log5

    I did:

    10^5 = (x^2 + 9) / x^2

    x's cancel out so I don't have any x to solve for. So clearly I messed up somewhere
    The x's don't cancel out you should have

    5=\frac{x^2+9}{x^2} \iff \frac{x^2+9}{x^2}-5=0 \iff \frac{x^2+9-5x^2}{x^2}=0 \iff \frac{9-4x^2}{x^2}=0

    This tells us that x cannot be equal to zero and now we need to set the quadratic in the numerator equal to zero and solve. Be sure to check your solutions in the original equaiton!
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  3. #3
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    Hello, Devi09!

    \text{Solve for }x \!:\;\;\log(x^2+9) - 2\log x\:=\:\log5

    Note that x must be positive.


    I avoid fractions when possible . . .

    We have: . \log(x^2+9) \:=\:\log 5+ 2\log x

    . . . . . . . . \log(x^2+9) \:=\:\log5 + \log(x^2)

    . . . . . . . . \log(x^2+9) \:=\:\log(5x^2)

    "Un-log": . . . . x^2 + 9 \:=\:5x^2

    And we have: . 4x^2 \:=\:9 \quad\Rightarrow\quad x^2 \:=\:\tfrac{9}{4} \quad\Rightarrow\quad x \:=\:\pm\tfrac{3}{2}

    Therefore: . x \:=\:\frac{3}{2}

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