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Math Help - Logarithm - Solve for X

  1. #1
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    Logarithm - Solve for X

    How would you go by solving these problems to find the value of x?

    1. (log_3 x)(log_4 3) = 2

    2. log x^3 - log y^2 =8
    log x^4 + log y^6 = 2
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by chrozer View Post
    How would you go by solving these problems to find the value of x?

    1. (log_3 x)(log_4 3) = 2

    2. log x^3 - log y^2 =8
    log x^4 + log y^6 = 2
    for one rewrite it as log_3(x)+\frac{1}{log_3(4)}...for the second one rewrite it as  log(x^4y^6)=2
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  3. #3
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    1. Know this identity: \log_{a} b = \frac{\log_{c} b}{\log_{c} a} for any workable c

    So:
    \log_{3}x \cdot \log_{4}{3} = 2
    \frac{\log x}{\log 3} \cdot \frac{\log 3}{\log 4} = 2

    You should be able to simplify it form here.

    2. Not sure where to go with MathStud's idea but I would just directly solve the system of equations:

    \begin{array}{ccccc} \log x^{3} & - & \log y^{2} & = & 8 \\ \log x^{4} & + &  \log y^{6} & = & 2 \end{array} . \quad \Rightarrow \quad . \begin{array}{ccccc} 3\log x & - & 2 \log y & = & 8 \\ 4\log x & + &  6\log y & = & 2 \end{array}

    \Rightarrow \begin{array}{cccccc} {\color{blue}9}\log x & - & {\color{blue}6}\log y & = & {\color{blue}24} & \mbox{(Multiplied by 3)} \\ 4\log x & + &  6\log y & = & 2 & \mbox{ } \end{array}

    Add them and solve for x.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    here is where I was going

    Quote Originally Posted by o_O View Post
    1. Know this identity: \log_{a} b = \frac{\log_{c} b}{\log_{c} a} for any workable c

    So:
    \log_{3}x \cdot \log_{4}{3} = 2
    \frac{\log x}{\log 3} \cdot \frac{\log 3}{\log 4} = 2

    You should be able to simplify it form here.

    2. Not sure where to go with MathStud's idea but I would just directly solve the system of equations:

    \begin{array}{ccccc} \log x^{3} & - & \log y^{2} & = & 8 \\ \log x^{4} & + & \log y^{6} & = & 2 \end{array} . \quad \Rightarrow \quad . \begin{array}{ccccc} 3\log x & - & 2 \log y & = & 8 \\ 4\log x & + & 6\log y & = & 2 \end{array}

    \Rightarrow \begin{array}{cccccc} {\color{blue}9}\log x & - & {\color{blue}6}\log y & = & {\color{blue}24} & \mbox{(Multiplied by 3)} \\ 4\log x & + & 6\log y & = & 2 & \mbox{ } \end{array}

    Add them and solve for x.
    \ln(x^4y^6)=2\Rightarrow{x^4y^6=e^2} and \frac{y^2}{x^3}=e^8 then solve...is that flawed logic?
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  5. #5
    o_O
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    Ah I see now, going with a direct comparison.

    And are we dealing with base 10 log or natural log?
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Haha

    Quote Originally Posted by o_O View Post
    Ah I see now, going with a direct comparison.

    And are we dealing with base 10 log or natural log?
    I never know anymore in math...I always taught that log(x)\equiv{log_{10}(x)} but in most mathematical text log(x)\equiv{ln(x)} so I just assume that I was taught wrong and use the latter...if he means log_{10}(x) then just replace the e with 10
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  7. #7
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    Quote Originally Posted by o_O View Post
    1. Know this identity: \log_{a} b = \frac{\log_{c} b}{\log_{c} a} for any workable c

    So:
    \log_{3}x \cdot \log_{4}{3} = 2
    \frac{\log x}{\log 3} \cdot \frac{\log 3}{\log 4} = 2

    You should be able to simplify it form here.

    2. Not sure where to go with MathStud's idea but I would just directly solve the system of equations:

    \begin{array}{ccccc} \log x^{3} & - & \log y^{2} & = & 8 \\ \log x^{4} & + &  \log y^{6} & = & 2 \end{array} . \quad \Rightarrow \quad . \begin{array}{ccccc} 3\log x & - & 2 \log y & = & 8 \\ 4\log x & + &  6\log y & = & 2 \end{array}

    \Rightarrow \begin{array}{cccccc} {\color{blue}9}\log x & - & {\color{blue}6}\log y & = & {\color{blue}24} & \mbox{(Multiplied by 3)} \\ 4\log x & + &  6\log y & = & 2 & \mbox{ } \end{array}

    Add them and solve for x.
    I solved for "x" and got 100, but when I plugged it back into the equation it does not equal the answer.

    EDIT - NVM....I'm an idiot. Plugged the "x" value for the "y" value also. But how would you solve for "y"? I can't get the same "y" value for both equation.

    MathStud - Thanx for the help...but I never learned that log(x) = ln(x)
    Last edited by chrozer; April 12th 2008 at 10:38 PM.
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Ok

    Quote Originally Posted by chrozer View Post
    I solved for "x" and got 100, but when I plugged it back into the equation it does not equal the answer.

    MathStud - Thanx for the help...but I never learned that log(x) = ln(x)
    well listing the adaptation I said we have \frac{x^3}{y^2}=10^8\Rightarrow{x^3=10^8y^2}...therefore using substitution we get x=100,y=\frac{1}{10}
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  9. #9
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    Quote Originally Posted by Mathstud28 View Post
    well listing the adaptation I said we have \frac{x^3}{y^2}=10^8\Rightarrow{x^3=10^8y^2}...therefore using substitution we get x=100,y=\frac{1}{10}
    I got that while trying to solve for both "x" and "y" but when I plug it back into the equation on my calculator...it does not equal to the answer of the equation.
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  10. #10
    o_O
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    \log (x^{3}) - \log (y^{2}) = \log\left(100^{3}\right) - \log \left(.1^{3}\right) = 6 - (-2) = 8

    \log (x^{4}) - \log (y^{6}) = \log\left(100^{4}\right) + \log \left(.1^{6}\right) = 8 + (-6) = 2
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  11. #11
    MHF Contributor Mathstud28's Avatar
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    You sure about that

    Quote Originally Posted by chrozer View Post
    I got that while trying to solve for both "x" and "y" but when I plug it back into the equation on my calculator...it does not equal to the answer of the equation.
    we have  log_{10}((10^2)^4)+log_{10}(10^{-6})=2\Rightarrow{8log_{10}(10)-6log_{10}(10)=2} \Rightarrow{2log_{10}(10)=2}\Rightarrow{2=2} and log_{10}((10^2)^3)-log_{10}(10^{-2})=8\Rightarrow{6log_{10}(10)+2log_{10}(10)=8} \Rightarrow{8log_{10}(10)=8}\Rightarrow{8=8}
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  12. #12
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    Quote Originally Posted by o_O View Post
    \log (x^{3}) - \log (y^{2}) = \log\left(100^{3}\right) - \log \left(.1^{3}\right) = 6 - (-2) = 8

    \log (x^{4}) - \log (y^{6}) = \log\left(100^{4}\right) + \log \left(.1^{6}\right) = 8 + (-6) = 2
    Ok thnx alot. I see what I did wrong. I put the exponent outside the parentheses. Thnx all for your help.
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