y=a^x, when a > 0, a <> 1: why a in this function could not be <0?

Question

Why does this certain function not include range less than 0?

Answer 1

This function could not be well defined for a < 0

Like some posters have mentioned with an integer x it is simple.

In fact with a rational it is sometimes simple.

Let x = p/q where gcd(p,q) = 1 then if q is odd. It is simple since a^(1/q) is well defined for q odd.

When x = p/q where gcd(p,q) = 1 and with q even then a^(1/q) is a complex number.

We run into another problem. There are q roots to each of the previous situations. Even if you pick the principal root then for the latter you still have a complex number.

So for y = f(x) = a^x with x in reals, restricting y to reals a must be positive.

I believe that even if you let y take complex values and if you take the principal root it wont be defined on most if not all irrational x.

Answer 2

You run into problems raising a negative base to a power that isn't an integer. (-1)^(1/2) means the square root of -1, which doesn't exist in the real number system. (-1)^2 can be done, but properties of exponents say this is the same as (-1)^(4/2) which involves that square root again. It just causes conflicts, so it's easier to leave it so that the base is positive.

Answer 3

Suppose a=-1. What would y be when x=1/2? This would be the square root of a negative number. This is only possible when you allow complex numbers.

Answer 4

It CAN be < 0, but in most algebra and calculus problems, you are dealing with a > 0

Answer 5

i thought u said "a > 0"... so it can't be less than zero since u difinted it that way

Answer 6

Just think about it...........
Let a= -0.5
Now try to evaluate it with different values of x.
If x is a whole integer it is easy, but what if x = 1.3 ?

Answer 7

a can't be less than zero because your problem said it was greater than zero