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  1. #1
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    Logs

    y = 6* 5^x and y = 9^x intersect at the point Q. Show that the x-coordinate of Q can be
    written as x = (1 + log3 2) / (2 − log3 5)
    Last edited by BabyMilo; April 24th 2010 at 07:40 AM.
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by BabyMilo View Post
    y = 6 × 5x and y = 9x intersect at the point Q. Show that the x-coordinate of Q can be
    written as x = (1 + log3 2) / (2 − log3 5)
    The question as written only has the solution (0,0)

    This is because we have 30x=9x.

    Are you sure you typed it correctly?
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  3. #3
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    Quote Originally Posted by e^(i*pi) View Post
    The question as written only has the solution (0,0)

    This is because we have 30x=9x.

    Are you sure you typed it correctly?
    sorry, corrected.
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  4. #4
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    and
    logy = log6 + xlog5 .....(1)
    logy = xlog9 ...........(2)
    So
    xlog9 =log6 +xlog5
    Solved for x and change the base to 3.
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  5. #5
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    Hello, BabyMilo!

    You really must learn to write exponential expressions . . .


    y \,=\, 6\cdot5^x and y \,=\, 9^x intersect at point Q.

    Show that the x-coordinate of Q is: .  x \;=\;\frac{1+\log_32}{2 - \log_35}

    We have: . 9^x \:=\:6\cdot5^x

    Take logs (base 3): . \log_3\left(9^x\right) \;=\;\log_3\left(6\cdot5^x\right)

    . . . . . . . . . . . . . . \log_3(3^2)^x \;=\;\log_3(6) + \log_3(5^x)

    . . . . . . . . . . . . . . \log_3(3^{2x}) \;=\;\log_3(3\cdot2) + \log_3(5^x)

    . . . . . . . . . . . . . . 2x\underbrace{\log_3(3)}_{\text{This is 1}} \;=\;\underbrace{\log_3(3)}_{\text{This is 1}} + \log_3(2) + x\log_3(5)

    . . . . . . . . . . . . . . . . . . 2x \;=\;1 + \log_3(2) + x\log_3(5)

    . . . . . . . . . . . 2x - x\log_3(5) \;=\;1 + \log_3(2)

    Factor: . . . . . x\bigg[2 - \log_3(5)\bigg] \;=\;1 + \log_3(2)

    Therefore: . . . . . . . . . . . x \;=\;\frac{1 + \log_3(2)}{2 - \log_3(5)}

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