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Math Help - Logarithm Equation

  1. #1
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    Exclamation Logarithm Equation

    Solve \lg(4^p - 4) - p\lg2 = \lg3
    I am supposed to do corrections for this question but I do not know how.
    The correct answer is p = 2
    Here are the steps I have done:
    \lg(4^p - 4) - p\lg2 = \lg3
    \lg(4^p - 4) - \lg2^p = \lg3<br />
    \lg\frac{4^p - 4}{2^p} = \lg3
    \frac{4^p - 4}{2^p} = 3
    4^p - 4 = 2^p \times 3
    \frac{2^{2p} - 4}{2^p} = 3
    \frac{2^{2p}}{2^p} - \frac{4}{2^p} = 3
    2^p - 2^{2 - p} = 3
    2^p(1 - 2^{2 - 1}) = 3
    2^p = -3

    Any help would really be appreciated, thank you very much in advance!
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  2. #2
    MHF Contributor Unknown008's Avatar
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    2^p - 2^{2 - p} = 3

    This line cannot become:

    2^p(1 - 2^{2 - 1}) = 3

    it becomes:

    2^p(1 - 2^{2 - 2p}) = 3



     2^{p} - \dfrac{4}{2^p} = 3

    From here, you can use a substitution of the form y = 2^p

    y - \dfrac{4}{y} = 3

    Now, can you continue?

    EDIT: Oops, sorry mr fantastic
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  3. #3
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    Quote Originally Posted by caramelcake View Post
    I am supposed to do corrections for this question but I do not know how.
    The correct answer is p = 2
    Here are the steps I have done:
    \lg(4^p - 4) - p\lg2 = \lg3
    \lg(4^p - 4) - \lg2^p = \lg3<br />
    \lg\frac{4^p - 4}{2^p} = \lg3
    \frac{4^p - 4}{2^p} = 3
    4^p - 4 = 2^p \times 3
    \frac{2^{2p} - 4}{2^p} = 3
    \frac{2^{2p}}{2^p} - \frac{4}{2^p} = 3
    2^p - 2^{2 - p} = 3
    2^p(1 - 2^{2 - 1}) = 3 Mr F says: This line is wrong. Expand it out to see why. 2^p 2^{1} \neq 2^{2-p}.
    2^p = -3

    Any help would really be appreciated, thank you very much in advance!
    4^p - 4 = 2^p \times 3

    \Rightarrow (2^{p})^2 - 3 \times 2^p - 4 = 0

    \Rightarrow (2^p - 4)(2^p + 1) = 0

    etc.
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  4. #4
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    Quote Originally Posted by caramelcake View Post
    I am supposed to do corrections for this question but I do not know how.
    The correct answer is p = 2
    Here are the steps I have done:
    \lg(4^p - 4) - p\lg2 = \lg3
    \lg(4^p - 4) - \lg2^p = \lg3<br />
    \lg\frac{4^p - 4}{2^p} = \lg3
    \frac{4^p - 4}{2^p} = 3
    4^p - 4 = 2^p \times 3
    4^p= (2^2)^p= (2^p)^2 so you can think of 4^p- 3(2^p)- 4= (2^p)^2- 3(2^p)- 4= 0 as a quadratic equation for 2^p. As Unknown008 (who is that masked man?) suggested, let y= 2^p and you get y^2- 3y- 4= 0. Solve that quadratic equation for y and then solve 2^p= y for p.

    \frac{2^{2p} - 4}{2^p} = 3
    \frac{2^{2p}}{2^p} - \frac{4}{2^p} = 3
    2^p - 2^{2 - p} = 3
    2^p(1 - 2^{2 - 1}) = 3
    2^p = -3

    Any help would really be appreciated, thank you very much in advance!
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  5. #5
    MHF Contributor Unknown008's Avatar
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    I'm not masked, I'm Jerry
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  6. #6
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    Thank you everybody! I am very touched by the number of responses to my question. Sorry for not replying earlier, I am not really feeling well today.
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  7. #7
    MHF Contributor Unknown008's Avatar
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    It's okay, we're glad to help
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  8. #8
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    Quote Originally Posted by Unknown008 View Post
    2^p - 2^{2 - p} = 3
    at this point either p or 2-p have to be 0, since the difference of 2 powers of 2 give an odd number only if the power of one is 0. also p cannot be zero since then 2-p=2 which gives -3.
    Last edited by Krahl; November 1st 2010 at 07:34 AM.
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