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Math Help - Solving a logarithmic equation with unlike bases.

  1. #1
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    Solving a logarithmic equation with unlike bases.

    Hi guys!

    Sorry to dump this one on you without any work done, but I'm really not sure where to start. I've only solved logarithmic equations with like bases so that the one-to-one property is used to eliminate the logarithms.

    Here is the equation:

    log_{4}(2x+1) = log_{2}(x-3)-log_{4}(x+5)

    The answer is given as:

    x = \frac{-5+\sqrt{41}}{2}

    Of course, if I'm going to learn anything, I need to see how the correct answer was arrived at.

    Thanks so much for the help!
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  2. #2
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    Quote Originally Posted by JennyFlowers View Post
    Here is the equation:
    \log_{4}(2x+1) = \log_{2}(x-3)-\log_{4}(x+5)
    The answer is given as:
    x = \frac{-5+\sqrt{41}}{2}
    That given answer is incorrect for the problem as written.
    That answer go with \log_{4}(2x+1) = \log_{2}(x{\color{blue}+}3)-\log_{4}(x+5)
    Note that \log_{2}(x+3)=2\log_{4}(x+3).

    Now it is all base four.
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  3. #3
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    Oops. That was my mistake, I put the wrong sign in the equation.

    Here is the correct problem:

    <br />
log_{4}(2x+1) = log_{2}(x+3)-log_{4}(x+5)<br />

    Could someone provide a step-by-step on how to reach the solution?

    Even with the conversion provided (thanks, Plato!) I'm not sure because I'd normally take the co-efficient and put it back as a power of the log. I'm really lost here.

    What's the best way to solve this one? Is there a way without doing that conversion?
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  4. #4
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    Combine to get \log_4\left(\frac{(x+3)^2}{(2x+1)(x+5)}\right)=0.

    That means  \frac{(x+3)^2}{(2x+1)(x+5)}=1
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  5. #5
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    Thanks, Plato.

    What property of logarithms allows you to do the conversion?

    <br />
\log_{2}(x+3)=2\log_{4}(x+3)<br />

    I'd like to know more about this property so I can understand exactly how to use it in the future.

    Thanks again!
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  6. #6
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    Also, I don't understand why:

    <br />
\log_4\left(\frac{(x+3)^2}{(2x+1)(x+5)}\right)=0<br />

    Simplifies to:

    <br />
\frac{(x+3)^2}{(2x+1)(x+5)}=1<br />

    Instead of:

    <br />
\frac{(x+3)^2}{(2x+1)(x+5)}=0<br />

    When you divide both sides by log_{4} shouldn't it give you a 0 on the right side instead of a 1?

    This does yield the correct answer for me though, so thanks again!
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  7. #7
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    If c^k=b all positive then k\log_b(N)=\log_c(N).

    Why 1 and not 0.
    Because 4^0=1.
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  8. #8
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    Does this logarithmic equation have a solution?

    Thank you!
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