Exponential function if a<0

• Mar 13th 2006, 05:04 AM
OReilly
Exponential function if a<0
I am learning about exponential function from math book for high-school.

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

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

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

Can someone explain me a little bit about that?
• Mar 13th 2006, 05:19 AM
earboth
Quote:

Originally Posted by OReilly
I am learning about exponential function from math book for high-school.

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

Hello,

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

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) $\displaystyle f \left( \frac{1}{2} \right)=(-2)^{\frac{1}{2}}=\sqrt{-2}$ which isn't a real number.

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

Greetings

EB
• Mar 13th 2006, 08:49 AM
OReilly
Quote:

Originally Posted by earboth
Hello,

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

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) $\displaystyle f \left( \frac{1}{2} \right)=(-2)^{\frac{1}{2}}=\sqrt{-2}$ 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 $\displaystyle 0^0$?
• Mar 13th 2006, 10:12 AM
TD!
Quote:

Originally Posted by OReilly
Just out of curiosity, how do you define $\displaystyle 0^0$?

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.
• Mar 13th 2006, 12:41 PM
ThePerfectHacker
Here let me demonstrate the graph $\displaystyle y=x^x$ for $\displaystyle x>0$
As you can see it is "reasonable" to say that at $\displaystyle x=0$ we have $\displaystyle y=0^0=1$ eventhough it is undefined there.
• Mar 13th 2006, 10:08 PM
rgep
To define logarithms of negative real numbers you have to go into the complex numbers. The fundamental formula is $\displaystyle e^{2\pi i} = 1$ which means that logarithms are really only defined up to the addition of a possible multiple of $\displaystyle 2 \pi i$. For positive reals, it makes sense to define "the" logarithm as a real number. For negative reals, we can say, for example, $\displaystyle \log (-x) = \log x + \pi i$ where the logarithm for the positive number x is the usual real value (this definition of log incorporates the famous formula $\displaystyle e^{\pi i} = -1$). But you could equally define it to be $\displaystyle \log x - \pi i$ or indeed $\displaystyle \log x + 1001\pi i$ if you wanted. You're always going to run up against the problem that now the formula $\displaystyle \log xy = \log x + \log y$ is only valid up to adding a possible correction of $\displaystyle 2\pi i$ or a multiple thereof (consider any sensible way of defining $\displaystyle \log (-1)^2$).