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Math Help - Log solve confusion

  1. #1
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    Log solve confusion

    Solving for x:

    log4(x+3)+log4(x-3)=2

    I'm doing something incorrect, I have gotten answers working through this as x=5 and x=5,-5

    I'm trying to understand why, and if in the end I would need to enter as a correct solution x=5 OR x=5,-5
    Last edited by gbrad88; November 30th 2012 at 10:42 AM.
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: Log solve confusion

    **EDIT For the record, the OP originally posted this:  \log_4(x+3)=\log_4(x-3)=2, which is what this response is based on **
    -----------------
    Do you actually mean this: one problem is \log_4(x+3) = 2, and a second problem is  \log_4(x-3) = 2? Because there is no way that  \log_4(x-3) = \log_4 (x+3)Assuming this is what you mean, then to solve for x remember the definition of a log: if \log_a b = c then  a^c = b. So for the first problem you have  4^2 = x+3 so x = 13. How do you get 5?
    Last edited by ebaines; November 30th 2012 at 10:46 AM.
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  3. #3
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    Re: Log solve confusion

    Quote Originally Posted by gbrad88 View Post
    Solving for x:l
    og4(x+3)=log4(x-3)=2
    I'm doing something incorrect, I have gotten answers working through this as x=5 and x=5,-5
    I'm trying to understand why, and if in the end I would need to enter as a correct solution x=5 OR x=5,-5

    I think that you mean \log_4(x+3)+\log_4(x-3)=2.

    That gives (x+3)(x-3)=4^2 or x^2-9=16.

    So x=\pm 5~.
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  4. #4
    MHF Contributor ebaines's Avatar
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    Re: Log solve confusion

    Quote Originally Posted by Plato View Post
    I think that you mean \log_4(x+3)+\log_4(x-3)=2.
    Good detective work!

    Quote Originally Posted by Plato View Post
    So x=\pm 5~.
    But if x = -5 then that means you have  \log_4 (-2) + \log_4 (-8) = 2, which isn't true. The answer is limited to x = +5.
    Last edited by ebaines; November 30th 2012 at 10:39 AM.
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  5. #5
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    Re: Log solve confusion

    ok, so my work that I did for:

    Solving for x:

    log4(x+3)+log4(x-3)=2

    would be 5,-5 or just 5?

    Sorry the response confused me?

    From reading, it seems it SHOULD be just "5"?
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  6. #6
    MHF Contributor ebaines's Avatar
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    Re: Log solve confusion

    gbrad88 - since you asked this in a pre-university algebra forum, can I assume that you have not covered complex numbers in your math classes? If that's so, then the logarithm of negative numbers is not something to consider, and for now you should limit your answers such that (x+3) and (x-3) are postive numbers only.
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  7. #7
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    Re: Log solve confusion

    My apologies. I just found this site and posted in the wrong place. Perhaps a mod can use it. I'm actually in a college algebra class. I fell a little behind due to taking multiple classes and a full work schedule and Im playing catch up now. Its not easy being an older guy that went back to college.

    So, looking at x = -5 and plugging it in then that means you something that isn't true. I think I understand it better now that I have written it out again. Correct me if Im wrong. THANK YOU for clearing this up.
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