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

[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.

and when you do... please do publish, my brain hurts already...

Thanks!

 

[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?

 

[hen]

it's not that easy...

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?

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..

 

[arildno]

It's not easy at all.

 

[Zurtex]

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

 

[joeboo]

First off, let's 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 existence
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.

 

[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, that's just simplifies back into the origional equation! This is hard.

 

[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.

 

[Zurtex]

First off, let's 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.

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

 

[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.

 

[Zurtex]

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.

Whoops, yeah, silly me haha.