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Solving of exponential equations and linear equations

  1. Apr 3, 2007 #1
    1. The problem statement, all variables and given/known data

    solve 3^x = 11-x
    2. Relevant equations

    I attempted by drawing both graphs but im searching for answers through algebra manipulation.

    3. The attempt at a solution

    x lg 3 = lg (11-x)
    x = [ lg (11-x) ]/[ lg 3 ]

    Any suggestions or solutions?:confused: Suggestions would be best..i would like to solve this on my own. " Show me the path not carry me through the journey " :biggrin:
  2. jcsd
  3. Apr 3, 2007 #2
    Are you allowed to graph them or are you trying to solve it algebracialy? If you have a graphing calculator you can just graph them and find the intersection.
  4. Apr 3, 2007 #3

    D H

    Staff: Mentor

    Note: This problem has an obvious exact solution.

    The solution form
    [tex]x = \frac{\ln(11-x)}{\ln 3}[/tex]
    suggests an iterator, which works quite nicely:
    [tex]x_{n+1} = \frac{\ln(11-x_n)}{\ln 3}[/tex]

    Doing something as simple as changing the 11 to 12 eliminates that obvious solution. The iterative approach still works nicely. Starting with [itex]x_0=0[/itex], the iterator [itex]x_{n+1} = \ln(12-x_n)/\ln 3}[/itex] converges to 2.08786968363036 in fourteen steps. Note that this is not the solution to the original problem.
  5. Apr 3, 2007 #4


    User Avatar
    Homework Helper

    You should notice that the LHS is a strictly increasing function (since (3x)' = 3x ln(3) > 0, for all x), and the RHS is a strictly decreasing function (since, (11 - x)' = -1 < 0). So if the equation does have solution, it can only have at most 1 solution. Do you see why? Hint: You can graph one increasing function, and one decreasing function to see if there is a chance that the two functions above intersect each other more than once.

    So, first thing is to guess the solution. Normally, the solution will be whole numbers.
    So, for x = 0, LHS = 1, RHS = 11, x = 0 is not the solution.
    x = 1, LHS = 3, RHS = 10, x = 1 is not the solution.
    x = 2, LHS = 9, RHS = 9, yay, x = 2 is one solution. :cool:
    Now, x = 2 is the solution.
    For any x > 2, since 3x is increasing, we have 3x > 32 = 9, and (11 - x) is decreasing, hence (11 - x) < (11 - 2) = 9
    So, for x > 2, we have: [tex]3 ^ x \neq 11 - x[/tex]

    You can do the same to show that:
    So, for x < 2, we have: [tex]3 ^ x \neq 11 - x[/tex]

    And hence, x = 2 is the only solution.

    Can you go from here? :)
    Last edited: Apr 3, 2007
  6. Apr 3, 2007 #5


    User Avatar
    Staff Emeritus
    Science Advisor

    Unscientific: There is no simple formula for solving problems when the unknown value appears both inside and outside a transcendental function.
    But note DH's statement "this problem has an obvious exact solution". "Trial and error" is a well respected mathematical method- as long as the "trials" don't take too long! Try some small integer values for x and see what happens.
  7. Apr 3, 2007 #6


    User Avatar
    Homework Helper

    Another approach is to re-arrange algebraically to the following form:

    y = f(x) = 0

    The solution is the zero of the equation. Plotting it easily reveals the solution.
    Last edited: Apr 3, 2007
  8. Apr 5, 2007 #7
    any other ways besides plotting the graph and guessing?
  9. Apr 5, 2007 #8

    D H

    Staff: Mentor

    You can estimate it numerically, which is going to be a lot more accurate than plotting. There are many techniques for finding the "roots" of a function. Wikipedia (http://en.wikipedia.org/wiki/Category:Root-finding_algorithms) and Mathworld (http://mathworld.wolfram.com/topics/Root-Finding.html) have extensive articles on several techniques.

    Another way to solve such problems is to find a way to express the function in the form x-g(x) = 0. This yields an iterator xn+1 = g(xn). The iterator may not converge to a solution. The iterator for this particular problem, xn+1 = log3(11-xn) works quite nicely for an initial guess between 0 and 10.
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