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How can I solve the following equation

  1. Mar 29, 2009 #1
    1. The problem statement, all variables and given/known data

    what is the procedure to solve the following equation for x?
    2^x + 6x = 16

    2. Relevant equations

    3. The attempt at a solution
    I tried as following but it didn't work,
    2^x + 6x = 16
    2^x + 6x = 2^4
    2^x + 6x = 2^4
    (2^x) / 2^4 + (6x)/ 2^4 = 1
    2^(x-4) + 6x/16 = 1
    2^(x-4) + 3x/8 = 1
    taking LCM,
    8[2^(x-4)] + 3x = 8
    [2^3][2^(x-4)] + 3x = 8
    2^(x-1) + 3x = 8
    after this stage I couldn't go farther..
    The answer is x=2, but how to obtain it?
  2. jcsd
  3. Mar 29, 2009 #2

    I doubt whether this is the sollution you wanted to see, but this kind of equations are usually solved numerically [please correct me if Im wrong].
    My proposition:

    we have a function:
    [tex]f(x) = 2^x + 6x - 2^4[/tex]
    three facts that are obvious:
    • the function is continuous and strictly increasing
    • [tex]f(0) < 0[/tex]
    • [tex]f(4) > 0[/tex].
    Realizing that it is a darboux function, we know that there exists such an "a", that [tex]f(a) = 0[/tex]. You can know write a computer program, to see the approximate number or you can guess it for yourself, cause the function you have is pretty student-friendly:

    [tex]f(0) < 0 \wedge f(4) > 0[/tex]
    [tex]f(1) < 0 \wedge f(4) > 0 [/tex]
    [tex]f(1) < 0 \wedge f(3) > 0 [/tex]
    [tex]f(2) = 0 [/tex]

    hey! thats my answer!

    maybe there is some smart way to do a few algebraic operations in order to get this '2' but I cant find it. I hope my post was somehow helpful to you.

  4. Mar 29, 2009 #3
    The way u showed is not what I wanted. I dont have any idea about darboux function.
    Can't we solve the above equation in any other easier way?

    And how do we solve x + ln x = c ( c is some constant) or, a^x + x = c (a,c r constants) equations?
  5. Mar 29, 2009 #4
    It can be solved graphically.

    Rearrange it:

    [itex]2^x + 6x = 16[/itex]

    [itex]2^x = 16 - 6x[/itex]

    Now plot both of the functions and it can be easily read from the graph (of course check it analytically) that they intersect at (2;4) so x = 2 is the root you're looking for.
  6. Mar 29, 2009 #5


    User Avatar
    Science Advisor

    If you are looking for a solution in terms of an "elementary function", there is none. That is generally true of equations that involve the variable both "inside" and "outside" a transcendental function.

    You could use the "Lambert W function",
    which is defined as the inverse function to f(x)= xex. For example, if x+ ln x= c, then ln x= x+ c. Taking the exponential of both sides, x= ex+ c= exec so, dividing both sides by ex, xe-x= ec. If we let u= -x, then x= -u and we have -ueu= ec or ueu= -ec. Applying the W function to both sides, u= W(-ec). Then x= -u= -W(-ec).
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