Solve for x in exponential function

[cshum00]

Homework Statement

So, i am given 3^x(2x) = 3^x + 2x + 1
And i want to solve for x.

Homework Equations

I only know that the solution is x=1 but i don't know how to get there.

The Attempt at a Solution

3^x(2x) = 3^x + 2x + 1
3^x(2x) - 3^x = 2x + 1
3^x(2x - 1) = 2x + 1
3^x = \frac{2x + 1}{2x - 1}
log(3^x) = log(\frac{2x + 1}{2x - 1})
3log(x) = log(2x + 1) - log(2x - 1)<- typo
xlog(3) = log(2x + 1) - log(2x - 1)
And then i am stuck.

 

[micromass]

I really doubt you can solve this analytically using elementary functions. Your best bet is guess and check to find the solutions.

 

[dirk_mec1]

This cannot be solved analytically.

 

[cshum00]

I am told by a Vietnamese friend that he solved it when he was in high school. Except that he no longer remembers how to do it. So, assuming that he did solve it, then it means that there is a analytical solution.

Edit: Ok, i am going to make a different approach.
3^x(2x) = 3^x + 2x + 1
3^x(2x) - 3^x - 2x = 1
3^x(2x - 1) - 2x = 1
3^x(2x - 1) - 2x + 1 = 1 + 1
3^x(2x - 1) - (2x - 1) = 2
(2x - 1)(3^x - 1) = 2

And now i am stuck...

 

[Mandelbroth]

Homework Statement

So, I am given 3^x(2x) = 3^x + 2x + 1
And I want to solve for x.

Homework Equations

I only know that the solution is x=1, but I don't know how to get there.

The Attempt at a Solution

3^x(2x) = 3^x + 2x + 1
3^x(2x) - 3^x = 2x + 1
3^x(2x - 1) = 2x + 1
3^x = \frac{2x + 1}{2x - 1}
log(3^x) = log(\frac{2x + 1}{2x - 1})
3log(x) = log(2x + 1) - log(2x - 1)
And then I am stuck.

Good job so far. Now, there are a couple ways you might find a solution for $x$. The one I'd think best fitting for this case would be graphing to find intersections. Or, you can look at your equation again. You know $1$ works. What other number works?

 

[cshum00]

Good job so far. Now, there are a couple ways you might find a solution for $x$. The one I'd think best fitting for this case would be graphing to find intersections. Or, you can look at your equation again. You know $1$ works. What other number works?

I thought of that too but let's stick with the analytical solution. And i just graphed the equations:
f_1(x) = 3^x
f_2(x) = \frac{2x + 1}{2x - 1}

The solutions are:
x=1 and x=-1

 

[Mentallic]

log(3^x) = log(\frac{2x + 1}{2x - 1})
3log(x) = log(2x + 1) - log(2x - 1)

The LHS is incorrect.

\log(3^x)=x\log(3)

I am told by a Vietnamese friend that he solved it when he was in high school. Except that he no longer remembers how to do it. So, assuming that he did solve it, then it means that there is a analytical solution.

That's a bold assumption to make considering even Wolfram Alpha wasn't able to give an analytic solution. Also, how is it possible that your friend remembered that he solved this exact problem long enough ago to simultaneously forget his solution?

 

[cshum00]

The LHS is incorrect.

\log(3^x)=x\log(3)

Thanks for noticing the typo.

That's a bold assumption to make considering even Wolfram Alpha wasn't able to give an analytic solution. Also, how is it possible that your friend remembered that he solved this exact problem long enough ago to simultaneously forget his solution?

I believe that there is a analytical solution for almost if not everything. It is just that we haven't discovered certain patterns or techniques to solver certain equations.

And software solutions follow certain set of rules. If specific rules used for solving certain forms of equations are not included, it simply doesn't know how to solve it.

Edit: Ok, here is another slightly approach
3^x(2x) = 3^x + 2x + 1
\frac{1}{2}3^x(2x) = \frac{1}{2}(3^x + 2x + 1)
3^x(x) = 3^x\frac{1}{2} + x + \frac{1}{2}
3^x(x) -3^x\frac{1}{2} - x = \frac{1}{2}
3^x(x - \frac{1}{2}) - x = \frac{1}{2}
3^x(x - \frac{1}{2}) - x + \frac{1}{2} = \frac{1}{2} + \frac{1}{2}
3^x(x - \frac{1}{2}) - (x - \frac{1}{2}) = 1
(x - \frac{1}{2})(3^x - 1) = 1

 

[micromass]

I believe that there is a analytical solution for almost if not everything. It is just that we haven't discovered certain patterns or techniques to solver certain equations.

And software solutions follow certain set of rules. If specific rules used for solving certain forms of equations are not included, it simply doesn't know how to solve it.

No, there are actually very few equations which have analytic solutions in function of elementary functions. If you want you can even proven when something has an elementary solution and when not. And I'm reasonably convinced that this does not have an analytic solution.

Also, certain types of software can tell you exactly when something is solvable or not. For example, if you ask wolfram alpha to solve an integral and he can't solve it, then you can be sure that there is no elementary solution. Why? Because the algorithm they use works if and only if there is an elementary solution. So if the algorithm fails, then there is no better algorithm that works.

It is largely the "fault" of HS textbooks which make it seem that everything is nice and solvable. But in reality, their exercises and examples deal with very special cases.

 

[Mentallic]

I believe that there is a analytical solution for almost if not everything. It is just that we haven't discovered certain patterns or techniques to solver certain equations.

An analytic solution has been defined to be an expression with certain characteristics, and many types of equations have been proven to have solutions that don't fit those certain characteristics.
You should read up on it.

And software solutions follow certain set of rules. If specific rules used for solving certain forms of equations are not included, it simply doesn't know how to solve it.

Of course, but when referring to certain well known and highly complex software such as wolfram alpha, I put more trust into its answer than high school solutions that have been lost in time.

But then again, I also don't believe that Fermat had a valid proof for Fermat's Last Theorem, regardless of how many margins he could have been given.

 

[cshum00]

No, there are actually very few equations which have analytic solutions in function of elementary functions. If you want you can even proven when something has an elementary solution and when not. And I'm reasonably convinced that this does not have an analytic solution.

Also, certain types of software can tell you exactly when something is solvable or not. For example, if you ask wolfram alpha to solve an integral and he can't solve it, then you can be sure that there is no elementary solution. Why? Because the algorithm they use works if and only if there is an elementary solution. So if the algorithm fails, then there is no better algorithm that works.

It is largely the "fault" of HS textbooks which make it seem that everything is nice and solvable. But in reality, their exercises and examples deal with very special cases.

I actually won't go and blame the HS textbooks. Actually it is the other way around. There is so much about the basics of math that we don't know. We only know limited patterns on how to solve certain things. For example, for addition/subtraction and multiplication/division, you find that there are so many different ways to solve them. However, as we go to more advanced levels like exponents and roots, we have so limited number of techniques to solve them. And it get worse as we progress to exponentials and logs.

I would say that there "is no better algorithm that works" at the current time. But it doesn't necessarily mean that there is no analytical solution at all.

 

[cshum00]

An analytic solution has been defined to be an expression with certain characteristics, and many types of equations have been proven to have solutions that don't fit those certain characteristics.
You should read up on it.


Of course, but when referring to certain well known and highly complex software such as wolfram alpha, I put more trust into its answer than high school solutions that have been lost in time.

But then again, I also don't believe that Fermat had a valid proof for Fermat's Last Theorem, regardless of how many margins he could have been given.

That is something very interesting we are touching on. I will definitely read on it

 

[micromass]

I actually won't go and blame the HS textbooks. Actually it is the other way around. There is so much about the basics of math that we don't know. We only know limited patterns on how to solve certain things. For example, for addition/subtraction and multiplication/division, you find that there are so many different ways to solve them. However, as we go to more advanced levels like exponents and roots, we have so limited number of techniques to solve them. And it get worse as we progress to exponentials and logs.

I would say that there "is no better algorithm that works" at the current time. But it doesn't necessarily mean that there is no analytical solution at all.

The point you're missing is that we can actually prove that no algorithm will work.
For example, if I ask you to solve $x^5 - x + 1 = 0$ using rational numbers, roots, and combinations, then it can be proven to be impossible. So no matter how we improve our algorithm it's impossible.

Another example closely related to your question. Given $x e^x = y$ for some constant $y$, then it can be proven that no elementary solution exists. Of course, for some special constants $y$ it will work, but not for all.

 

[cshum00]

The point you're missing is that we can actually prove that no algorithm will work.
For example, if I ask you to solve $x^5 - x + 1 = 0$ using rational numbers, roots, and combinations, then it can be proven to be impossible. So no matter how we improve our algorithm it's impossible.

Another example closely related to your question. Given $x e^x = y$ for some constant $y$, then it can be proven that no elementary solution exists. Of course, for some special constants $y$ it will work, but not for all.

Ok, can you elaborate on how you prove that there is no analytical solution?

 

[Ray Vickson]

Ok, can you elaborate on how you prove that there is no analytical solution?

Google "Galois Theory".

 

[STEMucator]

Ok, can you elaborate on how you prove that there is no analytical solution?

Have a look at this page, I'm sure it will enlighten you as to what micro is trying to get across :

http://en.wikipedia.org/wiki/Closed-form_expression

There are more instances that lack closed form solutions than ones which have closed form solutions.

 

[WannabeNewton]

I believe that there is a analytical solution for almost if not everything.

This is the complete, complete, complete, complete, complete, wait for it, complete opposite of reality.

 

[cshum00]

Have a look at this page, I'm sure it will enlighten you as to what micro is trying to get across :

http://en.wikipedia.org/wiki/Closed-form_expression

There are more instances that lack closed form solutions than ones which have closed form solutions.

That is the closest to what i am trying to say. I have always wondered if there are even more elementary arithmetic operations other than addition, subtraction, multiplication and division. I believe that the reason why we can't solve certain problems analytically is because we are missing certain elementary arithmetic operations that can define or co-define our elementary operations.

There is actually more base into it. It is like saying that we can't divide 1 over 3 and get a [strike]whole[/strike] [strike]rational[/strike] non-repeating decimal number. But that is because our number system is not divisible by three. If we had a ternary multiple number system, the answer divided by three would be a [strike]whole[/strike] [strike]rational[/strike] non-repeating decimal number.

 

[micromass]

That is the closest to what i am trying to say. I have always wondered if there are even more elementary arithmetic operations other than addition, subtraction, multiplication and division. I believe that the reason why we can't solve certain problems analytically is because we are missing certain elementary arithmetic operations that can define or co-define our elementary operations.

For sure, if we allow more operations, then we can solve more equations. For example, if you allow the Lambert W function as an elementary operation, then $y=xe^x$ is solvable. But no matter how many operations you introduce, there will always be equations that you can't solve.

 

[D H]

But no matter how many operations you introduce, there will always be equations that you can't solve.

That you can't solve analytically.

This is why numerical methods are so important.

 

[cshum00]

@micromass
The question becomes how do you know that for all the set of infinity operations and infinity equations, there is a subset of equations which becomes unsolvable no matter the operations used?

@D H
Numerical solutions are great but what i hate about numerical solutions is that they are "estimation" of the actual result. Although we get an accurate to whatever digit we make to compute, it is not the "actual" answer. Except for some cases when we round it up and it turns out to be the exact answer but it is not always the case.

 

[Ray Vickson]

@micromass
The question becomes how do you know that for all the set of infinity operations and infinity equations, there is a subset of equations which becomes unsolvable no matter the operations used?

@D H
Numerical solutions are great but what i hate about numerical solutions is that they are "estimation" of the actual result. Although we get an accurate to whatever digit we make to compute, it is not the "actual" answer. Except for some cases when we round it up and it turns out to be the exact answer but it is not always the case.

You want things that are not achievable---this has been proven. If you insist on rejecting numerical solutions you will not be able to practice in the technical world. Besides, there is really not much difference between a "numerical" solution and a numerically-evaluated exact solution to a real-world problem. For example, consider the equation $x^4 - 3x^3 + 5x^2 - 4x - 7= 0.$ There are formulas for the roots; one of the two real roots is given by
3/4-1/12*3^(1/2)*((-13*(3844+36*53741^(1/2))^(1/3)+2*(3844+36*53741^(1/2))^(2/3)-760)/(3844+36*53741^(1/2))^(1/3))^(1/2)+1/12*((-78*(3844+36*53741^(1/2))^(1/3)*((-13*(3844+36*53741^(1/2))^(1/3)+2*(3844+36*53741^(1/2))^(2/3)-760)/(3844+36*53741^(1/2))^(1/3))^(1/2)-6*((-13*(3844+36*53741^(1/2))^(1/3)+2*(3844+36*53741^(1/2))^(2/3)-760)/(3844+36*53741^(1/2))^(1/3))^(1/2)*(3844+36*53741^(1/2))^(2/3)+2280*((-13*(3844+36*53741^(1/2))^(1/3)+2*(3844+36*53741^(1/2))^(2/3)-760)/(3844+36*53741^(1/2))^(1/3))^(1/2)+18*3^(1/2)*(3844+36*53741^(1/2))^(1/3))/(3844+36*53741^(1/2))^(1/3)/((-13*(3844+36*53741^(1/2))^(1/3)+2*(3844+36*53741^(1/2))^(2/3)-760)/(3844+36*53741^(1/2))^(1/3))^(1/2))^(1/2)
If you needed a numerical answer, you could evaluate this formula to get 2.206912782 or 2.2069115688725810670 or 2.2069115688725810642745469101658853710256141267195, or... . However, in a real-world problem (this IS, after all, Physics Forums) you might not need a value better than about 2.207. Getting 10--15 digit accuracy in a purely numerical method is nowadays no problem at all.

 

[cshum00]

@Ray Vickson
Things that are not achievable today doesn't mean they aren't achievable tomorrow.
Things that seem to be correct today may be proven wrong tomorrow.

I am not rejecting numerical solutions, i just don't like it because it is not elegant.
It is just a personal opinion. However, i do understand that at the moment, there are
things that can only be don with numerical methods.

Let's jut keep the conversation here. I see that there is no point of going back and forth.
What everyone is saying is pretty much that my opening question can't be answered
with current analytical methods.

 

[STEMucator]

That is the closest to what i am trying to say. I have always wondered if there are even more elementary arithmetic operations other than addition, subtraction, multiplication and division. I believe that the reason why we can't solve certain problems analytically is because we are missing certain elementary arithmetic operations that can define or co-define our elementary operations.

There is actually more base into it. It is like saying that we can't divide 1 over 3 and get a whole number. But that is because our number system is not divisible by three. If we had a ternary multiple number system, the answer divided by three would be a whole number.

While what everyone else is saying is indeed correct, I find your view interesting as it challenges the current consensus.

I've observed that mathematics is taught in systematic manner without much consideration beyond what has been quantified and proven ( Blatent Formalism ). The analytical solutions you would seek to a problem depend on the rules of the number system you happen to be using.

The problem with that I find is depending on how those rules are defined, you can only analytically solve a very small amount of problems within the confines of those rules. The rules also happen to be very strict ( precise if you will ) and as a result, limit capabilities.

I wonder if there's a universal number system perhaps? One which we don't have the ability to abstract quite now, but would be able to solve any equation analytically without the need of numerical assistance. I wonder how the rules would be defined and how the structure of the actual numbers would behave.

Not in our lifetime I don't think :(.

 

[Ray Vickson]

That is the closest to what i am trying to say. I have always wondered if there are even more elementary arithmetic operations other than addition, subtraction, multiplication and division. I believe that the reason why we can't solve certain problems analytically is because we are missing certain elementary arithmetic operations that can define or co-define our elementary operations.

There is actually more base into it. It is like saying that we can't divide 1 over 3 and get a whole number. But that is because our number system is not divisible by three. If we had a ternary multiple number system, the answer divided by three would be a whole number.

I guess I mis-understood your original question. You gave an equation, and you seemed to be asking how you could solve it now, today. Apparently, though, you wanted to know how you might, possibly, solve it at some unspecified, hypothetical time perhaps far in the future.

 

[cshum00]

@Zondrina
Well, the thing is that the more i look back when i review my basics, the more i question them. I even ask myself how come 1+1=2? I know it is something very intuitive but i know there is more into it that just that. I also know that there is a proof on that but i was never able to find it. Maybe someone knows the source and refer it to me.

Another thing is like the +/- signs. The general rule says that a multiplication of same signs will always give you a + sign. And that the multiplication of opposite signs will always give you a - sign. But that is true until we stumble into imaginary numbers. I know we can debate that imaginary numbers are not "real" numbers or have no real representation in the real world. But is that really true? For me, we can't really say until we know all about it.

@Ray Vickson
Well, i did want to solve my original question with our current "analytical" methods. Like i said before, my Vietnamese friend told me that he solved that problem in high school. That is of course, if and only if he was not lying. There is also the possibility that he was just bragging about it.

 

[dirk_mec1]

This discussion is getting out of hand, if I understand correctly the OPs question is answered, right?