Maths

slove : x^2=4^x

asked by Rabin kaji
  1. x^2=4^x
    let's use logs
    log (x^2) = log (4^x)
    2 log x = x log 4 , not getting anywhere, can't isolate x

    graphing it to approximate ....
    http://www.wolframalpha.com/input/?i=x%5E2%3D4%5Ex
    shows a solution at x = appr -.64

    How about Newton's method?
    let y = x^2 - 4^x
    dy/dx = 2x - (ln4)(4^x)

    new x = x - (x^2 - 4^x)/(2x - ln4(4^x))
    start with x = -.64
    x ... newx
    -.64 -.641186
    -.641186 -.641185744
    -.641185744 -.641185744, wow just 3 steps!!

    check:
    -.641185744^2 = .411119159
    4^-.641185744 = .41119159

    x = -.641185744

    posted by Reiny
  2. Just to chime in, in case you haven't studied calculus yet, there's always the bisection method. Let
    f(x) = x^2 - 4^x
    You want to find where f(x) = 0. Since
    f(-1) > 0
    f(0) < 0
    f(x) = 0 for some x in the interval [-1,0]
    so, start guessing at the center of the interval. pick whichever half of the interval the change of sign appears in.
    f(-0.75) = 0.2089
    f(-0.625) = -0.2982
    f(-0.6875) = 0.0871
    f(-0.65625) = 0.2804
    ... 9 steps later ...
    f(-0.6412353515625) = 0.000091888

    As you can see, it converges rather more slowly than Newton's method. It took 13 iterations to get the same accuracy of 3 steps with Newton's method.

    posted by Steve

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