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Math Help - More logs

  1. #1
    Member smmmc's Avatar
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    More logs

    1)log3(x^2-3x-1)=0

    2)-logx(25)=-2
    =x^-2=25

    i know the base is 5 but confused whether it is negative or positive.

    thanks
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  2. #2
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    Quote Originally Posted by smmmc View Post
    1)log3(x^2-3x-1)=0
    \log_b 1 = 0 for b > 0, so
    \log_3 (x^2-3x-1) = 0 becomes
    x^2-3x-1 = 1.
    Solve the quadratic:
    \begin{aligned}<br />
x^2-3x-2 &= 0 \\<br />
x &= \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-2)}}{2(1)} \\<br />
x &= \frac{3 \pm \sqrt{9 + 8}}{2} \\<br />
x &= \frac{3 \pm \sqrt{17}}{2}<br />
\end{aligned}


    2)-logx(25)=-2
    =x^-2=25

    i know the base is 5 but confused whether it is negative or positive.
    -\log_x 25 = -2
    First get rid of the negative signs:
    \log_x 25 = 2.
    As you said, x = 5 (and it's positive).


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    Last edited by yeongil; July 18th 2009 at 03:50 AM.
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  3. #3
    Member smmmc's Avatar
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    b>1 always apply ? what about when b<1 ?

    thanks heaps
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  4. #4
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    Oops, typo. I meant b > 0 and b ≠ 1. Of course, in logarithms you can't have a negative base (b < 0).


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    Last edited by yeongil; July 18th 2009 at 05:26 AM. Reason: because I'm sleep deprived. :(
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  5. #5
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    e^(i*pi)'s Avatar
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    You can't have a base of 1 either

    b>0 \: , b \neq 1
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  6. #6
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    If the original problem had been
    log_x(25)= -2
    then it would have been 25= x^{-2}= \frac{1}{x^2} so that x^2= \frac{1}{25} and x= \frac{1}{5}, still positive.
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