Hard and tricky

1. Jun 5, 2005

hen

Hello!
at first i thought this one is very easy:

3^(x/2)+1=2^x

the question is: find x, analiticly. meaning, solve it as if you didn't know that x=2, also don't prove x=2 is the only answer, or that it is an answer. don't guess the answer... find it analiticly.

Thanks!

2. Jun 5, 2005

quetzalcoatl9

hey now, don't make us do all the work!

it would be easier if the x's were all down below, and not exponents, wouldn't it?

so how do you get the x's out of the exponents?

3. Jun 5, 2005

hen

it's not that easy...

i tried all kinds of ln, tried doubled integrals, to try maybe it will look familier with something, nothing worked so far.
i don't think it's easy. asked 2 phd, and they didn't have a quick answer (on of them phd in math).
i really set on it hard, and didn't figure out, if the solution is easy, and "was there all the time", i'll be dissapointed..

4. Jun 5, 2005

arildno

It's not easy at all.

5. Jun 5, 2005

Zurtex

I don't beleive there is a way to solve this in an algebraic way.

6. Jun 5, 2005

joeboo

First off, lets do $x = 2t$, and set $f(t) = 4^t - 3^t - 1$. This, in my opinion, just makes the whole thing easier to work with algebraically. It is not a necessary change, but does make the steps easier.

Hints for proving there is a unique zero of f(t):
f(t) is clearly continuous, therefore it satisfies intermediate value theorem. This can be used to show existance
For uniqueness, take the derivative of f(t). Certain properties of logarithm and the exponential function come into serious play here.

With regards to "analytically finding the zero", this is really a difficult notion. I look at f(t) and say, "by inspection, it's zero is at t = 1". However, that's not really an analytic method. Therefore, I'd suggest using "newton's method" to approximate the zero ( giving it sufficient bounds ), and then simply saying "Oh that's close to t = 1 ( or x = 2 ), let's see what f(1) is equal to!" This is about as close as ( I can imagine ) you will get to an "analytic solution". It is possible that newton's method will converge quite rapidly and, given suitable initial points, may converge exactly to t = 1. Don't take my word on this, as I haven't tried it.

I'm presuming you are familiar with derivatives and newton's method as this is in calculus forum. If you are vaguely familiar with newton's method, try looking it up on wikipedia or the wolfram mathworld site for a refresher.

Hope this helps.

7. Jun 5, 2005

Cyrus

I managed to reduce it to this, if anyone can solve this you'll have your anwser:

$$3^x = 4^x - 2(3^{x/2})-1$$

Edit: Rats, thats just simplifies back into the origional equation! This is hard.

Last edited: Jun 5, 2005
8. Jun 6, 2005

Curious3141

It isn't "hard", it's just one of those things that's impossible to solve in a standard analytic fashion, that's all. Even equations with trivially simple roots like this one can be impossible to solve exactly unless one just guesses and substitutes.

I would reserve the term "hard" for problems where there is a solution but it's difficult (but not impossible) to find.

9. Jun 6, 2005

Zurtex

1 is a root of this function, 1 is not a root of the orriginal function given...

10. Jun 6, 2005

matt grime

And? t=1 is a solution of this form , and thus x=2 of the original as was noted.

It can be shown that any solution must be positive, and that for positive t, or x, that it mustbe unique since the function is monotonically increasing for x or t greater than zero (the derivative is positive). Thus there is a unique real solution. Obviously by insepction t=1 is a root and thus it is the only real one. Of course, this hardly counts as "analytic" in the sense intended, I imagine.

11. Jun 6, 2005

Zurtex

Whoops, yeah, silly me haha.