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Solve for x in exponential function

  1. Jul 23, 2013 #1
    1. The problem statement, all variables and given/known data
    So, i am given [tex]3^x(2x) = 3^x + 2x + 1[/tex]
    And i want to solve for x.

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

    3. The attempt at a solution
    [tex]3^x(2x) = 3^x + 2x + 1[/tex]
    [tex]3^x(2x) - 3^x = 2x + 1[/tex]
    [tex]3^x(2x - 1) = 2x + 1[/tex]
    [tex]3^x = \frac{2x + 1}{2x - 1}[/tex]
    [tex]log(3^x) = log(\frac{2x + 1}{2x - 1})[/tex]
    [tex]3log(x) = log(2x + 1) - log(2x - 1)<- typo[/tex]
    [tex]xlog(3) = log(2x + 1) - log(2x - 1)[/tex]
    And then i am stuck.
    Last edited: Jul 23, 2013
  2. jcsd
  3. Jul 23, 2013 #2
    I really doubt you can solve this analytically using elementary functions. Your best bet is guess and check to find the solutions.
  4. Jul 23, 2013 #3
    This cannot be solved analytically.
  5. Jul 23, 2013 #4
    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.
    [tex]3^x(2x) = 3^x + 2x + 1[/tex]
    [tex]3^x(2x) - 3^x - 2x = 1[/tex]
    [tex]3^x(2x - 1) - 2x = 1[/tex]
    [tex]3^x(2x - 1) - 2x + 1 = 1 + 1[/tex]
    [tex]3^x(2x - 1) - (2x - 1) = 2[/tex]
    [tex](2x - 1)(3^x - 1) = 2[/tex]

    And now i am stuck...
    Last edited: Jul 23, 2013
  6. Jul 23, 2013 #5
    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?
  7. Jul 23, 2013 #6
    I thought of that too but let's stick with the analytical solution. And i just graphed the equations:
    [tex]f_1(x) = 3^x[/tex]
    [tex]f_2(x) = \frac{2x + 1}{2x - 1}[/tex]

    The solutions are:
    x=1 and x=-1
  8. Jul 23, 2013 #7


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    The LHS is incorrect.


    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?
  9. Jul 23, 2013 #8
    Thanks for noticing the typo.

    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
    [tex]3^x(2x) = 3^x + 2x + 1[/tex]
    [tex]\frac{1}{2}3^x(2x) = \frac{1}{2}(3^x + 2x + 1)[/tex]
    [tex]3^x(x) = 3^x\frac{1}{2} + x + \frac{1}{2}[/tex]
    [tex]3^x(x) -3^x\frac{1}{2} - x = \frac{1}{2}[/tex]
    [tex]3^x(x - \frac{1}{2}) - x = \frac{1}{2}[/tex]
    [tex]3^x(x - \frac{1}{2}) - x + \frac{1}{2} = \frac{1}{2} + \frac{1}{2}[/tex]
    [tex]3^x(x - \frac{1}{2}) - (x - \frac{1}{2}) = 1[/tex]
    [tex](x - \frac{1}{2})(3^x - 1) = 1[/tex]
    Last edited: Jul 23, 2013
  10. Jul 23, 2013 #9
    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.
  11. Jul 23, 2013 #10


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    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.
  12. Jul 23, 2013 #11
    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.
  13. Jul 23, 2013 #12
    That is something very interesting we are touching on. I will definitely read on it
  14. Jul 23, 2013 #13
    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.
  15. Jul 23, 2013 #14
    Ok, can you elaborate on how you prove that there is no analytical solution?
  16. Jul 23, 2013 #15

    Ray Vickson

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    Google "Galois Theory".
  17. Jul 23, 2013 #16


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


    There are more instances that lack closed form solutions than ones which have closed form solutions.
  18. Jul 23, 2013 #17


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    This is the complete, complete, complete, complete, complete, wait for it, complete opposite of reality.
  19. Jul 23, 2013 #18
    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.
    Last edited: Jul 23, 2013
  20. Jul 23, 2013 #19
    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.
  21. Jul 23, 2013 #20

    D H

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    That you can't solve analytically.

    This is why numerical methods are so important.
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