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Can anyone solve this equation?

  1. Jan 14, 2004 #1
    My teacher told us in class that the only way we can solve this equation is by using numerical methods (ie. graphing calculator). I was wondering if anyone would be able to solve it mathematically. I tried plugging it in to Mathematica but it couldn't solve it.

    [tex]5^x + 9^x = 92[/tex]
  2. jcsd
  3. Jan 14, 2004 #2


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    I would say that solving numerically is solving mathematically! Perhaps you meant "algebraically" or "analytically". I suspect you could use the Lambert W function but you might not consider that any "better" than doing it numerically.
  4. Jan 14, 2004 #3
    This is just a wild guess, but can it be solved by using logs?
  5. Jan 14, 2004 #4
    I tried, but I don't think so.

    HallsOfIvy: Well, you know what I mean. I'd like to have a non-infinite mathematical expression for x. I suppose it's probably impossible to do that. I'll look up the Lambert W function.

    If it helps, an equivalent problem is the following:

    Given [itex]f(x) = 5^x + 9^x[/itex], find [itex]f^{-1}(x)[/itex].

    I wonder if there is an expression for [itex]f^{-1}(x)[/itex] in the form of an infinite series.
  6. Jan 14, 2004 #5
    I looked up the Lambert W function. Funny, it's similar to what I just said above about the inverse function. Well, I'll see if I can get it to work.
  7. Jan 15, 2004 #6
    Hello to you.

    Hi there. Well, I'm at school right now... if I have time, I'll
    show my answer. =) I hope you don't need it at a particular time.
  8. Jan 16, 2004 #7


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    Your teacher is right, but I think the distinction between analytic and numeric solutions is not so solid as you might think. Consider the equation


    One tends to think that this is solvable in closed form as [tex]\inline{x=\sin^{-1}{.6}}[/tex]. But why? because most calculators have an arcsin button? If you were to do it by hand, you would be hard-pressed to find a good numerical approximation. Conversely, it would be no harder to build a calculator with a (admittedly useless) [tex]\inline{f^{-1}}[/tex] button where [itex]f(x) = 5^x + 9^x[/itex] than to build one with an arcsin button. How? By Newton's method: Consider the function

    [tex]F(x)=x-(5^x+9^x){\log(5^x+9^x)-\log(92)\over \log(5)5^x+\log(9)9^x}[/tex]

    Starting with a first guess say x=2, each application of [itex]F(x)[/itex] gives a better approximation to the solution. For example [itex]F(F(2))[/itex] already gives the solution to 11 places. Apply F once more and you get the solution to 22 places.

    (Edit: it's way faster yet if you first take the log of both sides of the original equation.)
    Last edited: Jan 16, 2004
  9. Jan 16, 2004 #8
    Yep, I understand what you mean. Actually, I went through the same kind of thought process as you. I thought about the sine function. I was just hoping there would be a nice solution to the problem in terms of usual functions (not weird ones like the inverse of 5^x + 9^x), but I suppose there isn't. I wonder if that franz32 guy actually got anywhere.

    I suppose that's the key: finding solutions in terms of ordinary functions. For example, the quintic equation can only be solved using the inverse of the function x^5 + x (well, that's one way to solve it).

    So now I can reformulate my original question: Is there any solution to that equation in terms of ordinary functions? And I suppose the answer is no.
  10. Jan 18, 2004 #9
    Sorry for waiting...

    Hello there. Honestly, I found out that this is not as simple as it is. solving this analytically may be difficult. if you know calculus, it will be easy if you are going to use Newton's method (which I haven't encountered yet in my university). If you are not familiar with calculus, then you will need to use some
    root-finding method such as the Bisection Method, the Secant Method,
    or the False Position Method. I believe that "krab" has the point. =)

    First of all, when you say "linear interpolation" for finding roots, there are actually two methods that use linear interpolation: "False Position" (or, in the Latin wording, "Regula Falsi"), and the Secant Method. False Position always brackets the root. That is, at each stage, the method produces two numbers a_n and b_n, so that the root r is between them: a_n < r < b_n. At the next stage, it replaces one of the numbers a_n and b_n with a new value and leaves the other value unchanged, and the new values a_(n+1) and b_(n+1) still bracket the root. The Secant Method produces a sequence of numbers that do not necessarily bracket the root.

    Your first criterion might be more useful for the False Position
    method, but you are correct that for certain functions, it will tend
    to be only one endpoint that moves, and the method will converge very
    slowly, by that criterion. The second and third criteria would be more appropriate for the Secant Method, but again, there are functions for which the second criterion will produce very slow convergence.

    The third criterion can be useful, but it depends upon whether you are primarily interested in a value that makes the function small, or
    whether you really need the correct value of the (exact) zero. If the
    root is a multiple one, so that the function is roughly quadratic or
    cubic in (x-r), where r is the exact root, then the function will be
    very flat near the root. For example, x might be .01 units from r, but f(x)~10^(-6), so relatively poor approximations of the root r will produce very small function values.

    The best one to use depends upon what you are trying to find: a good approximation to the root, or just a value x which produces a small f(x).

    Well, that's it. Sorry, I thought it's an easy one... I lately realized that this involves numerical analysis, which I'm going to take up 2 years from now.
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