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Math Help - Some math problems i cant seem to figure out?

  1. #1
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    Unhappy Some math problems i cant seem to figure out?

    9. Solve the equation and express the solution in exact form: log (x+3)=1-logx


    10. Solve for indicated variable: P=40,000 X e^t/3 , for t



    11. Use any method ( analytic or graphical) to solve the equation. If necessary, round the answer to the nearest thousandth: e^x+ln10=6e^x
    Last edited by mr fantastic; January 5th 2009 at 10:39 PM.
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  2. #2
    Member Last_Singularity's Avatar
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    Quote Originally Posted by rotw121 View Post
    9. Solve the equation and express the solution in exact form: log (x+3)=1-logx


    10. Solve for indicated variable: P=40,000 X e^t/3 , for t



    11. Use any method ( analytic or graphical) to solve the equation. If necessary, round the answer to the nearest thousandth: e^x+ln10=6e^x

    (9) Assuming that the log is base-10, raise 10 to both sides of the equation to cancel out the logs:
    10^{log(x+3)} = 10^{1-log(x)}

    x+3 = (10^1)(10^{-log(x)})

    x+3 = (10)(1/x)

    x+3 = 10/x

    Can you take it from there?


    (b) Take the natural logarithm for both sides:
    P=40000 e^{t/3}

    ln(P) = ln(40000) + ln(e^{t/3})

    ln(P) = ln(200^2) + t/3

    ln(P) = 2 ln(200) + t/3

    Then solve for t from here.


    (c) Once again since there is a sort of a e^x or similar expression, take the natural log of both sides
    e^x+ln10=6e^x

    ln10 = 5e^x

    \frac{1}{5} ln10 = e^x

    ln(\frac{1}{5} ln10) = x

    QED
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  3. #3
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    Quote Originally Posted by rotw121 View Post
    9. Solve the equation and express the solution in exact form: log (x+3)=1-logx


    10. Solve for indicated variable: P=40,000 X e^t/3 , for t



    11. Use any method ( analytic or graphical) to solve the equation. If necessary, round the answer to the nearest thousandth: e^x+ln10=6e^x
    to #9:

    I assume that \log(x)=\log_{10}(x) . The domaion of this equation is d = \mathbb{R}_0^+

    \log_{10} (x+3)=1-\log_{10}(x)~\implies~x+3 = \dfrac{10}x ~\implies~x^2+3x-10=0~\implies~x=-5~\vee~x=2

    Since -5\notin d the only solution is x = 2

    to #10:

    P=40,000 \cdot e^{\frac t3}~\implies~\dfrac P{40000} = e^{\frac t3}~\implies~\dfrac t3=\ln\left(\dfrac P{40000} \right)

    Therefore t=\ln(P^3)-\ln(40000^3)

    to #11:

    If there isn't any typo the result is x=\ln\left(\dfrac15 \ln(10) \right)

    If and only if you mean:

    e^{x+\ln(10)}=6e^x~\implies~10e^x=6e^x~\implies~10  =6 which is a rather seldom!
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