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Math Help - How to solve this inequality?

  1. #1
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    How to solve this inequality?

    Is there a way to solve for x in the equation: 5^(x-2) > -1?
    It is 5 to the power of x-2.

    I tried to take log on both sides but log a negative number doesn't seem right.
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  2. #2
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    That is true for all x.

    For all t, 5^t>0.
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  3. #3
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    Ignore this - elementary mistake - see below
    Last edited by e^(i*pi); October 10th 2010 at 08:51 AM. Reason: putting complex number into form z=a+bi
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    Quote Originally Posted by e^(i*pi) View Post
    There are no real solutions because 5^(x-2) \geq 0 \: , \: x \in \mathbb{R}
    In the complex plane the principle solution of \ln(-1) \text{  is  } i\pi (this follows from Euler's indentity - which happens to be my username)
    (x-2)\ln(5) > i\pi
    x > \dfrac{2}{\ln(5)}+\dfrac{\pi}{\ln(5)}\,i
    What are you saying?
    The inequality is true for every real number.

    Moreover there is no ordering in the complex numbers.
    So what does x > \dfrac{2}{\ln(5)}+\dfrac{\pi}{\ln(5)}\,i mean?
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  5. #5
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    Quote Originally Posted by Plato View Post
    What are you saying?
    The inequality is true for every real number.

    Moreover there is no ordering in the complex numbers.
    So what does x > \dfrac{2}{\ln(5)}+\dfrac{\pi}{\ln(5)}\,i mean?
    No idea how I missed that >.<
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  6. #6
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    Quote Originally Posted by Plato View Post
    That is true for all x.

    For all t, 5^t>0.
    ah thanks! how come I didn't think of that...haha...
    thanks a lot!
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