What number to the power of itself equals 2 its something like 1.5596.... How is it calculated?

Answer 1

In order to answer this question, you need to know about natural logarithms and implicit differentiation as well as the Newton-Raphson (N-R) method for numerical approximation.

Let y = x^x

Take natural logarithms to give ln(y) = ln(x^x) = x*ln(x)

Now differentiate: (1/y)*(dy/dx) = 1*ln(x) + x*1/x = ln(x) + 1

Therefore, dy/dx = y*[ln(x) + 1] = x^x*[(ln(x) + 1]

Now, solving for x^x = 2 is the same as solving for the root of x^x - 2 = 0. So let f(x) = x^x - 2 and use the N-R method to solve for its root.

Start with an estimate x(0) = 2, say.

Use the iteration x(n+1) = x(n) - f(x(n))/f'(x(n)) for n = 1, 2, 3, ...

This gives x(4) = 1.559610563... and x(5) = 1.559610469 : sufficient accuracy for most people.