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Math Help - Does this logarithmic equation have a solution?

  1. #1
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    Does this logarithmic equation have a solution?

    I must solve the following logarithmic equation:

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

    I found the answer as:

    x = \frac{-17\pm\sqrt{305}}{2}

    Which gives me two approximate answers:

    0.232 and -17.232

    The negative answer clearly cannot be considered a solution. But what about the positive answer? When I sub it back into the original equation, it appears to cause a problem at log_{2}(x-3) since it would give this log a negative argument. However, since it can be converted to log_{4}(x-3)^2 does that mean that 0.232 does work as a solution?

    Thank you!
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by JennyFlowers View Post
    I must solve the following logarithmic equation:

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

    I found the answer as:

    x = \frac{-17\pm\sqrt{305}}{2}

    Which gives me two approximate answers:

    0.232 and -17.232

    The negative answer clearly cannot be considered a solution. But what about the positive answer? When I sub it back into the original equation, it appears to cause a problem at log_{2}(x-3) since it would give this log a negative argument. However, since it can be converted to log_{4}(x-3)^2 does that mean that 0.232 does work as a solution?

    Thank you!
    The only solution is  x = \frac{-17 - \sqrt{305}}{2}
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  3. #3
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    Quote Originally Posted by JennyFlowers View Post
    I must solve the following logarithmic equation:

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

    I found the answer as:

    x = \frac{-17\pm\sqrt{305}}{2}

    Which gives me two approximate answers:

    0.232 and -17.232

    The negative answer clearly cannot be considered a solution. But what about the positive answer? When I sub it back into the original equation, it appears to cause a problem at log_{2}(x-3) since it would give this log a negative argument. However, since it can be converted to log_{4}(x-3)^2 does that mean that 0.232 does work as a solution?


    No. The argument of the real logarithm to any base must be positive, no matter if you raise to the second or to the 14th power, and this means your equation has no solution

    Tonio


    Thank you!
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  4. #4
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    Quote Originally Posted by harish21 View Post
    The only solution is  x = \frac{-17 - \sqrt{305}}{2}
    But that gives an answer of -17.232, which isn't a solution.
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  5. #5
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by JennyFlowers View Post
    But that gives an answer of -17.232, which isn't a solution.
    oops... sorry.. I completely ignored the negative value. The system wont have a solution.
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