1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    Hi,

    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.

    rahl
     
  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

    HallsofIvy

    User Avatar
    Staff Emeritus
    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",
    http://mathworld.wolfram.com/LambertW-Function.html,
    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).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook