I'm not understanding the question. If f(x)= one thing, the answer is this. The other gives you that, but what exactly is the question? This is all they give you:

If f(x)=
&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp e², x < ln2
&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp 2, x >/= ln2

Today has turned out to be a horrible day. I'm going to go jump into bed and hope for a better tomorrow.

<font color=lime>You posted twice... ;) </font>

You must be tired - you double posted. [img]graemlins/cool.gif[/img]

Quote from other thread:

[ 01-17-2004, 07:18 AM: Message edited by: GokuZool ]

I think it's just my computer doing its best to try to annoy me. Took me ten minutes just to load IW this morning.

OK, this I know. How am I to decide what x is?

Nothing is stated besides this question. It's a multiple choice question, so I suppose I could include those:
&nbsp&nbspa. ½
&nbsp&nbspb. ln2
&nbsp&nbspc. 2
&nbsp&nbspd. e²
&nbsp&nbspe. It is nonexistent

Are you SURE the original problem's typed correctly? Right now, as is, it's just a discontinuous piecewise function, with a horizontal line at about 7.4 until it reaches about .7 on the x axis...it stays as 2 there after. So it's just two horizontal lines crammed together, without even matching, if that helps any (which I doubt it does). Right now, though, the question is about like saying "If Billy has four apples then:"

a)8
b)16
c)radical 19
d)see what I'm trying to get at here?

If you've provided EVERYTHING, then the question has a typo.

I think I might have stumbled upon the solution, but I don't know how explain it on paper. One of the definitions of an exponential function includes:

e^x=y, if lny=x

So, if x=ln2, then:

e^x, x < ln2
2, x >/= ln2

Both would come out to equal the same thing. But that is only if you take x to be equal to ln2 on both accounts. Then the answer would be 2 (c).

THE EXPONENTIAL FUNCTION OF EULER'S NUMBER REARS IT'S UGLY HEAD! Sorry for misunderstanding, 10^-9 byte...That 2 in the original kinda threw me...Ok, now it's a continuous piecewise function, that comes togeth nicely at about .7, which is kinda what I figured...You should be all right with 2 as your answer.

Well, depending on typos you have two different situations:

based on you first post
f(x)=| e^2 for all x less than the natural log of 2

OR
| 2 for all x greater than or equal too ln(2).

f(x) is one of two constants.

based of the second post:
f(x)=|e^x ...... this is a variable function until x>= ln(2), then f(x) = 2

The other stuff you posted is just to show that the inverse function of ln(x) = e^x thats all don't confuse these concepts.

There are two types of inverses. Multiplicitave and Additive.

For multiplicative, x times its inverse = 1
For additive, x plus it's inverse = 0

For example 2 x ¹/₂ = 1
2 + (-2) = 0

I understand this. But under the circumstances, I have to prove one of those answers to be true.

I don't get what you're saying, Night Stalker. If I solve ln2, it comes out to be .6931471806. If I plug this number into my e^x, I get out 2. That is what I meant in my previous post.

[ 01-17-2004, 04:06 PM: Message edited by: Nanobyte ]

<span style="color: lightblue">There's a very good reason for that: ln x is equivalent to log_ex ... and from the definition of a log:
If log_xy = z , then x^z = y .

[ 01-18-2004, 04:11 AM: Message edited by: LennonCook ]