# Thread: Exponential function if a<0

1. ## 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?

2. 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

3. 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 $\displaystyle 0^0$?

4. 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.

5. 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.

6. 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 $\displaystyle \log (-1)^2$).