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Math Help - Two Logarithmic Function Problems

  1. #1
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    Two Logarithmic Function Problems

    I'm having trouble solving these two problems:

    1.) log_2(x+7) + log_2(x) = 3

    2.) e^x + 2 = 8e^{-x}

    Your assistance is much appreciated.
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  2. #2
    Super Member wingless's Avatar
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    Ignore the solution to the first question, there's a mistake in it.
    log_c a + log_c b = log_c a.b
    So,
    log_2(x+7) + log_2(x) = 3
    log_2{x^2 + 7} = 3
    x^2 + 7 = 2^3 = 8
    x^2 = 1
    x={-1,1}
    Don't forget to check the results because x+7 > 0 and x>0.
    Ah, x = -1 doesn't satisfy x>0. So we remove it from the solution set.
    x=1




    ----------------------------------------------




    e^x + 2 = 8e^{-x}

    e^x + 2 = \frac{8}{e^x}

    Let a=e^x

    a + 2 = \frac{8}{a}

    Multiply both sides by a,

    a^2 + 2a = 8

    a^2 + 2a - 8 = 0

    (a+4)(a-2)=0

    a=\{ -4 , 2 \}


    Put these a values back in a=e^x,

    -4 = e^x

    x = \ln -4 (Not real!)
    (We exclude this because it's impossible to have such real x)


    2 = e^x

    x = \ln 2
    Last edited by wingless; March 15th 2008 at 03:06 PM.
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  3. #3
    o_O
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    The rule should be: \log_{c}a + \log_{c}b = log_{c}(ab)

    \log_{2}(x + 7) + \log_{2}(x) = 3
    \log_{2}\left[(x + 7)(x)\right] = 3
    \log_{2}\left(x^{2} + 7x\right) = 3
    x^{2} + 7x = 2^{3}
    x^{2} + 7x = 8

    Thus, a simple quadratic.
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  4. #4
    Super Member wingless's Avatar
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    Sorry, o_O is right. Ignore my solution to the first problem, I didn't multiply 7 by x. But the rest is similar, find the roots, check them for x+7>0 and x>0.
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