# Trying to determine domain of f(t)=4.5e^t

1. Nov 6, 2007

### poohbear1986

heres a question im trying to figure out if this equations domain
f(t)=4.5e^t where e is 2.1783

im thinking that the answer is all positve numbers. because it can't equal zero so all negative numbers cause it to be 0

2. Nov 6, 2007

### Diffy

I used to think of domains like this (pre-calculus).

The Domain of a function are the numbers that you are allowed to plug in for your variable (in your case 't').

There are 2 golden rules that you can not break with real functions.

You can not divide by 0.

You can not take the square root of a negative number.

If you do either of these two things you are no longer dealing with real numbers.

So, in your function, are you breaking either of these two rules? If not, then your domain is all real numbers. If you find numbers that break either of those two rules, then your domain is all real numbers, except those numbers which break the rules.

3. Nov 6, 2007

### poohbear1986

so i can have a negative exponet, and it all still will be real numbers

4. Nov 6, 2007

### SnipedYou

Yes because the law of negative exponents states:

$$a^{-1} = \frac{1}{a}$$

5. Nov 6, 2007

### poohbear1986

thank you i apprecite the help immensely

6. Nov 7, 2007

### matt grime

I don't think you have said enough. If I replaced e by pi, then the domain is not so obvious: raising numbers to irrational powers can only be done by saying x^n=exp(n logx). (exp(y)is e^y)

Your reasoning is therefore only going to be circular unless you use the power series definition of e^x.

7. Nov 7, 2007

### HallsofIvy

It looks like you are confusing "domain" with "range". "Domain" is the set of possible x-values while "range" is the set of y-values. You are certainly correct that e to any power is positive and so the range of this function is "all positive real numbers".

8. Nov 7, 2007

### Diffy

More generally,

$$a^{-n} = \frac{1}{a^{n}}$$