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Math Help - Exponential function if a<0

  1. #1
    Senior Member OReilly's Avatar
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    Exponential function if a<0

    I am learning about exponential function from math book for high-school.

    I have learned about y = a^x if a>0 and x \in R. I have understand pretty much of it, but I didn't find any explanation if a<0 (why?).

    I have tried to understand by myself properties of exponential function y = a^x if a<0 and x \in Z.

    This is what I think are properties:
    1) for a<0 and x =2k function is defined but there is no inverse function
    2) for a<0 and x =2k+1 function is defined but there is inverse function

    Can someone explain me a little bit about that?
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  2. #2
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    Quote Originally Posted by OReilly
    I am learning about exponential function from math book for high-school.

    I have learned about y = a^x if a>0 and x \in R. I have understand pretty much of it, but I didn't find any explanation if a<0 (why?).
    Hello,

    I'll try to explain by an example:
    Take f(x)=y=(-2)^x

    a) f(1)=-2 and f(2)=4. That means between x=1 and x=2 the graph of the "function" has to cross the x-axis. But you'll never get the result 0 (zero), if the base isn't zero too.
    b) f \left( \frac{1}{2} \right)=(-2)^{\frac{1}{2}}=\sqrt{-2} which isn't a real number.

    To avoid all these difficulties exponential functions are only defined for positive bases except 1.

    Greetings

    EB
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    Senior Member OReilly's Avatar
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    Quote Originally Posted by earboth
    Hello,

    I'll try to explain by an example:
    Take f(x)=y=(-2)^x

    a) f(1)=-2 and f(2)=4. That means between x=1 and x=2 the graph of the "function" has to cross the x-axis. But you'll never get the result 0 (zero), if the base isn't zero too.
    b) f \left( \frac{1}{2} \right)=(-2)^{\frac{1}{2}}=\sqrt{-2} which isn't a real number.

    To avoid all these difficulties exponential functions are only defined for positive bases except 1.

    Greetings

    EB
    Yes, I understand now.
    Thanks!

    Just out of curiosity, how do you define 0^0?
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  4. #4
    TD!
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    Quote Originally Posted by OReilly
    Just out of curiosity, how do you define 0^0?
    It is generally left undefined. When one defines it though, mostly for the purpose of simplyfing sum- and product notation and power series, it is usually defined as 1. The reason why this is more logical than choosing it to be 0 is because of the real limit of x^x for x going to zero is 1, you may also want to check the graph of the function y = x^x for x tending to 0.
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  5. #5
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    Here let me demonstrate the graph y=x^x for x>0
    As you can see it is "reasonable" to say that at x=0 we have y=0^0=1 eventhough it is undefined there.
    Attached Thumbnails Attached Thumbnails Exponential function if a&lt;0-picture9.gif  
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  6. #6
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    To define logarithms of negative real numbers you have to go into the complex numbers. The fundamental formula is e^{2\pi i} = 1 which means that logarithms are really only defined up to the addition of a possible multiple of 2 \pi i. For positive reals, it makes sense to define "the" logarithm as a real number. For negative reals, we can say, for example, \log (-x) = \log x + \pi i where the logarithm for the positive number x is the usual real value (this definition of log incorporates the famous formula e^{\pi i} = -1). But you could equally define it to be \log x - \pi i or indeed \log x + 1001\pi i if you wanted. You're always going to run up against the problem that now the formula \log xy = \log x + \log y is only valid up to adding a possible correction of 2\pi i or a multiple thereof (consider any sensible way of defining \log (-1)^2).
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