Hi, i just wanted to clarify some things I'm not sure about. Is it that e^x cannot equal zero or must be greater than zero or both? Also, would anybody be able to explain why?
Thanks!
The value of e
x, for any real number x, will always be greater than zero (and cannot be equal to zero). One way you can confirm this is by simply sketching a graph of y=e
x, where you will notice that the graph is always above the x-axis (y=0), but approaches y=0 (i.e. gets closer and closer to 0, but never really is zero).
But how do we explain this conceptually? Well, it's first important to take note that e is really just a positive number, 2.72 (to 2 d.p.). The second thing to realize is that we are just applying index laws to this arbitrary positive number, where the power applied is x. I am going to apply 4 different cases to hopefully help you understand why the above explanation is the case. The first and simplest case is when x>=1. In this case, the value of e
x will either be e (when x=1, since a number to the power of 1 is the number itself) OR it will be greater than e. You can see this for yourself by plugging in values of x. e
2 is approximately 7.39, e
1.5 is 4.48, and you could go on and on just plugging in values of x which are greater than 1 for you to see that they will always be greater than e, and thus
greater than zero.
The second case is 0 < x < 1. In this case, you will notice that the value of e
x will actually be lower than the value of e since we are applying a power which is less than one. Once again, you can try plugging in random values of x between 0 and 1 to notice this. However, at x=0 (our third case), the value of e
x is
1 since virtually all numbers to the power of 0 are 0 (we don't talk about 0
0 lol). Thus, returning to our second case, we can see that the value of e
x will be between 1 and e when x is between 0 and 1. In other words, for both the second and third cases, e
x is
greater than zero.
Our fourth and final case is when x<0, when it is a negative number. Basically, as x gets lower and lower, the value of e
x will become closer and closer to zero. We can remember this due to our index law for negative exponents: a
-b = 1/a
b.
Now in this case, our negative number is x. For example, let x=-1. Then, e
-1 = 1/e. Keep in mind that this is a positive number divided by a positive number, so it cannot be a negative number. This applies for any negative value of x. Furthermore, note that the only possible way for this to equate to zero is if the numerator is equal to zero, which is not possible here. That explains why it will still always be
greater than zero.
But why does it approach zero as x becomes greater? Essentially, it's because you're dividing 1 by a larger number as x decreases further. To illustrate this, compare x=-1 (e
-1, or 1/e
1), x=-2 (e
-2, or 1/e
2), and x=-3 (e
-3, or 1/e
3). You can see that as x becomes more negative, we are dividing by a larger number which makes the value really close to zero, but never zero.
Hopefully this all makes sense
Edit: beat by Jinju-san, but I'll still post this in case it's any help haha