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Math Help - help solving an equation involving logs

  1. #1
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    help solving an equation involving logs

    \sqrt{log(x) - 3} = log(x) - 3

    I'm not sure how to solve this. Logically, it would seem to me that log(x) - 3 must equal 0 or 1 for \sqrt{log(x) - 3} to equal log(x) - 3. I'm not sure how to prove it though. Could someone point me in the right direction?
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  2. #2
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    Quote Originally Posted by absvalue View Post
    \sqrt{log(x) - 3} = log(x) - 3

    I'm not sure how to solve this. Logically, it would seem to me that log(x) - 3 must equal 0 or 1 for \sqrt{log(x) - 3} to equal log(x) - 3. I'm not sure how to prove it though. Could someone point me in the right direction?
    HI

    You can let log x = b

    \sqrt{b-3}=b-3 .. square both sides and with some algebra

    b^2-7b+12=0

    (b-4)(b-3)=0

    b=4 or b=3

    Then log x =4 or log x =3

    Solve for x .
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  3. #3
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    Thanks! That makes sense. So...

    \log(x) = 4 \Longleftrightarrow 10^4 = x

    or

    \log(x) = 3 \Longleftrightarrow 10^3 = x
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  4. #4
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    Quote Originally Posted by absvalue View Post
    Thanks! That makes sense. So...

    \log(x) = 4 \Longleftrightarrow 10^4 = x

    or

    \log(x) = 3 \Longleftrightarrow 10^3 = x
    Absolutely
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  5. #5
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    Quote Originally Posted by absvalue View Post
    \sqrt{log(x) - 3} = log(x) - 3

    I'm not sure how to solve this. Logically, it would seem to me that log(x) - 3 must equal 0 or 1 for \sqrt{log(x) - 3} to equal log(x) - 3. I'm not sure how to prove it though. Could someone point me in the right direction?
    Here is another attempt:

    If z = \sqrt{\log(x)-3}

    then your equation becomes:

    z = z^2~\implies~z^2-z=0~\implies~z(z-1)=0~\implies~ z = 0~\vee~z=1

    Now re-substitute to calculate x.
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