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Thread: Logs/Exponent Questions

  1. #1
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    Logs/Exponent Questions

    A few question i cannot solve and need some help on:
    1)$\displaystyle ln(5x)-ln(3-2x)=1$
    2)$\displaystyle ln(x)+ln(3x+1)=1$
    3)$\displaystyle 8e^-x-e^x$


    P.S
    Last edited by Paymemoney; Dec 7th 2009 at 03:41 PM. Reason: meant to be log to the base e
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Paymemoney View Post
    A few question i cannot solve and need some help on:
    1)$\displaystyle log(5x)-log(3-2x)=1$
    2)$\displaystyle log(x)+log(3x+1)=1$
    3)$\displaystyle 8e^-x-e^x$


    P.S
    I'll do the first as an example

    $\displaystyle log(5x)-log(3-2x)=1$

    $\displaystyle log\left(\frac{5x}{3-2x}\right) = 1$

    $\displaystyle \frac{5x}{3-2x} = 10^1 = 10$

    $\displaystyle 5x = 30-20x$

    $\displaystyle 25x = 30$

    $\displaystyle x = \frac{6}{5}$
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  3. #3
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    ok, well I've solved 1) & 2) but how would i solve 3)?
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  4. #4
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Paymemoney View Post
    ok, well I've solved 1) & 2) but how would i solve 3)?
    To solve 3 it will need to be equal to something

    As it stands you can only take out a factor of $\displaystyle e^{-x}$

    $\displaystyle 8e^{-x}-e^x = e^{-x}(8-e^{2x})$
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  5. #5
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    i think the equation is meant to be $\displaystyle 8e^{-x}-e^{x}=2$
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  6. #6
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    Quote Originally Posted by Paymemoney View Post
    i think the equation is meant to be $\displaystyle 8e^{-x}-e^{x}=2$
    In that case multiply throughout by $\displaystyle e^x$

    $\displaystyle 8-e^{2x}=2e^x$

    Rearrange into the form $\displaystyle ax^2+bx+c=0$

    $\displaystyle e^{2x}+2e^x-8=0$

    Solve the quadratic (this one factorises) and then find x using the natural log. Remember that one solution will not be real.

    I get solutions of

    $\displaystyle x = ln(2)$ REAL

    and

    $\displaystyle x = ln(-4) = ln(4) + i\pi$ COMPLEX


    You may only need the REAL solution
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  7. #7
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    ok thanks
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