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# Easy logarithm question but im stupid watch

1. (Original post by Mr Laminator)
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
2. (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

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
3. 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.
4. (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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