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