solve for x from 2logx=x/25

Question

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Answer 1

Question: solve for x from 2logx=x/25.

1st) 2.log(x) = log(x^2)

2nd) log_b(a) = c means, by definition, that a = b^c

3rd) so, log_10(x^2) = (x/25) means that x^2 = 10^(x/25)

Use the Newton_Raphson method to solve this

Let f(x) = x^2 - 10^(x/25)

Then f'(x) = 2x - (1/25)*10^(x/25)*ln(10)

Finding a starting value for x0

x=1:   f(x) = 1 - 10^(0.04) = 1 - 1.1 = -0.1

x = 2:  f(x) = 4 - 10^(0.08) = 4 - 1.2 = 2.8

There is a change of sign in f(x) from -0.1 to +2.8 between x = 1 and x = 2.

So the root lies in this range. Start the iterations with x0 = 1.

The iteration is:

x_(n+1) = x_n - f(x_n) / f'(x_n), i.e.

x_(n+1) = x_n - {x_n^2 - 10^(x_n/25)} /{2x_n - (1/25)*10^(x_n/25)*ln(10)}

n   x_n                      f(x_n)                        f'(x_n)                 x_(n+1)=x_n - f(x_n) / f'(x_n)

0    1                     -0.96478196e-1        1.899010626        1.050804453

1    1.050804453   0.002569069            2.000145869        1.049520012

2    1.049520012   0.000001643           1.997588989         1.049519190

3    1.049519190

To 5 dp, we have the solution as: x3 = 1.04952