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Easy logarithm question but im stupid watch

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    (Original post by Mr Laminator)
    Please would u be able to post your numerical solution?
    Here's one method, which is marginally less tiresome than some of the alternatives...

    Newton-Raphson Iteration:

    xn+1 = xn - f(xn)/f'(xn)

    with f(x) = 5x - 2x - 1
    and f '(x) = 5x .ln5 - 2x .ln2

    Starting with x1=1

    x2= 1-(2/6.66) = 0.6997
    x3= 0.6997 - (0.4597/3.837) = 0.58
    x4= 0.58 - (0.048/3.057) = 0.564
    x5= 0.564 - (0.00073/2.96) = 0.564

    So x=0.56 (2dp)

    Check shows
    f(0.555) is negative
    f(0.565) is positive
    so correct to 2dp

    Useful P4 practice!

    Aitch
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    (Original post by Spenceman_)
    2x + 1 = 5x

    ln(2x + 1) = ln5x

    ln(2x + 1) = xln5


    x = [ln(2x + 1)]/ln5

    Setting up an iterative process:-

    xn+1 = [ln(2xn + 1)]/ln5

    Start with x1 = 1

    Iteration converges to x = 0.563895524
    Quite encouraging that we get to the same destination via different routes!

    Linear Interpolation or Interval Bisection anyone?

    Aitch
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    Cheers Spenceman and Aitch, I vaguely understand what you've done. I haven't done Newton-Raphson or iterations or anything yet, I'm still at the 'C1-C2: shove a number in until you find one that fits' stage.
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    (Original post by Mr Laminator)
    Cheers Spenceman and Aitch, I vaguely understand what you've done. I haven't done Newton-Raphson or iterations or anything yet, I'm still at the 'C1-C2: shove a number in until you find one that fits' stage.
    The Iteration Formula process and Newton-Raphson Iterations are OK.

    The fact that few TSR Mathematicians seem to be rushing to perform Interval Bisection or Linear Interpolation says a lot for these two procedures...

    Aitch
 
 
 
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