2+2. OK? That's nice, what about it?

[Nanobyte]

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.

[LordKathen]

You posted twice...

[GokuZool]

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

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

Quote from other thread:

Gabrielles blades: if im not mistaken, it means that if x is less than ln2 then y is equal to e^2 and if x is greater than or equal to ln2 then y is equal to 2 course, there is no basic question per say when you state an equation, does your book for this particular round of equations state something like If X is this (for example, 2), then what is y? If so you would be able to 'solve' it. If not, the question may be simply asking you to explain the properties of the equation as i have posited above.

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

[Nanobyte]

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.

Originally posted by Gabrielles blades: if im not mistaken, it means that if x is less than ln2 then y is equal to e^2 and if x is greater than or equal to ln2 then y is equal to 2

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

course, there is no basic question per say when you state an equation, does your book for this particular round of equations state something like If X is this (for example, 2), then what is y? If so you would be able to 'solve' it.

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

[Ilander]

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.

[Nanobyte]

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).

[Ilander]

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.

[Night Stalker]

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

[Nanobyte]

Originally posted by Night Stalker: 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.

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

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 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 ]

[LennonCook]

Originally posted by Nanobyte: 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 ex, I get out 2. That is what I meant in my previous post.

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 ]