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Math Help - Logging out soon ( it's still about logs)

  1. #1
    Member M670's Avatar
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    Logging out soon ( it's still about logs)

    Solve for :

    Can I not rewirte this as x=4/7(log6(sqrt108(2048))/(2/sqrt3)^3(9)^3)
    then i solve it an get 5.874

    I am sure its x=4/7(log6(100077696)
    Last edited by M670; October 13th 2012 at 03:13 PM.
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  2. #2
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    Re: Logging out soon ( it's still about logs)

    Quote Originally Posted by M670 View Post
    Solve for :

    Can I not rewirte this as x=4/7(log6(sqrt108(2048))/(2/sqrt3)^3(9)^3)
    then i solve it an get 5.874
    I am sure its x=4/7(log6(100077696)
    Here is a hint.
    \log_6(\sqrt{108})=0.5(\log_6(6^2\cdot 3)=0.5(2+\log_6(3))
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  3. #3
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    Re: Logging out soon ( it's still about logs)

    Hello, M670!

    Solve for x\!:\;x \;=\;\tfrac{4}{7}\left[\log_6\sqrt{108} - 3\log_6\left(\tfrac{2}{\sqrt{3}}\right) + 3\log_6(9) + \log_6(2028)\right]

    Note that:

    \log_6\sqrt{108} \;=\;\log_6 (2^2\cdot3^3)^{\frac{1}{2}} \;=\;\log_6\left(2\cdot3^{\frac{2}{3}}\right)

    -3\log\left(\tfrac{2}{\sqrt{3}}\right) \;=\;-3\log_6\left(\tfrac{2}{3^{\frac{1}{2}}}\right) \;=\; -\log_6\left(\tfrac{2}{3^{\frac{1}{2}}}\right)^3 \;=\;-\log_6\left(\tfrac{2^3}{3^{\frac{3}{2}}}\right)
    . . . . . . . . . =\;-\left[\log_6(2^3) - \log_6(3^{\frac{3}{2}})\right] \;=\;\log_6(3^{\frac{3}{2}}) - \log_6(2^3)

    3\log_6(9) \;=\; \log_6(9^3) \;=\; \log_6(3^2)^3 \;=\; \log_6(3^6)

    \log_6(2048) \;=\;\log_6(2^{11})


    Then we have:

    x \;=\;\tfrac{4}{7}\left[\log_6(2\cdot3^{\frac{3}{2}}) + \log_6(3^{\frac{3}{2}}) - \log_6(2^3) + \log_6(3^6) + \log_6(2^{11})\right]

    x \;=\;\tfrac{4}{7}\log_6\left[\frac{2\cdot3^{\frac{3}{2}}\cdot3^{\frac{3}{2}}}{2  ^3}\cdot 3^6\cdot 2^{11}\right]

    x \;=\;\tfrac{4}{7}\log_6(2^9\cdot3^9)

    x \;=\;\tfrac{4}{7}\log_6(2\cdot3)^9

    x \;=\;\tfrac{4}{7}\cdot9\log_6(2\cdot3)

    x \;=\;\tfrac{36}{7}\underbrace{\log_6(6)}_{ \text{This is 1}}

    x \;=\;\frac{36}{7}
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  4. #4
    Member M670's Avatar
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    Re: Logging out soon ( it's still about logs)

    Quote Originally Posted by Soroban View Post
    Hello, M670!


    Note that:

    \log_6\sqrt{108} \;=\;\log_6 (2^2\cdot3^3)^{\frac{1}{2}} \;=\;\log_6\left(2\cdot3^{\frac{2}{3}}\right)

    -3\log\left(\tfrac{2}{\sqrt{3}}\right) \;=\;-3\log_6\left(\tfrac{2}{3^{\frac{1}{2}}}\right) \;=\; -\log_6\left(\tfrac{2}{3^{\frac{1}{2}}}\right)^3 \;=\;-\log_6\left(\tfrac{2^3}{3^{\frac{3}{2}}}\right)
    . . . . . . . . . =\;-\left[\log_6(2^3) - \log_6(3^{\frac{3}{2}})\right] \;=\;\log_6(3^{\frac{3}{2}}) - \log_6(2^3)

    3\log_6(9) \;=\; \log_6(9^3) \;=\; \log_6(3^2)^3 \;=\; \log_6(3^6)

    \log_6(2048) \;=\;\log_6(2^{11})


    Then we have:

    x \;=\;\tfrac{4}{7}\left[\log_6(2\cdot3^{\frac{3}{2}}) + \log_6(3^{\frac{3}{2}}) - \log_6(2^3) + \log_6(3^6) + \log_6(2^{11})\right]

    x \;=\;\tfrac{4}{7}\log_6\left[\frac{2\cdot3^{\frac{3}{2}}\cdot3^{\frac{3}{2}}}{2  ^3}\cdot 3^6\cdot 2^{11}\right]

    x \;=\;\tfrac{4}{7}\log_6(2^9\cdot3^9)

    x \;=\;\tfrac{4}{7}\log_6(2\cdot3)^9

    x \;=\;\tfrac{4}{7}\cdot9\log_6(2\cdot3)

    x \;=\;\tfrac{36}{7}\underbrace{\log_6(6)}_{ \text{This is 1}}

    x \;=\;\frac{36}{7}
    Wow almost every rule is being used in the problem
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