Find the value of x in the equation (√(x-1/x))^x^2 = (1/2)^(x+1)

Question

I want the process to arrive at x

Answer 1

Take logs of each side:

½x²log(x-1/x)=-(1+x)log(2),

½x²(log(x²-1)-log(x))+(1+x)log(2)=0.

Let f(x)=½x²(log(x²-1)-log(x))+(1+x)log(2). Define log(x) as ln(x), because any log base is valid and because ln(x), etc., is easy to differentiate.

f'(x)=xln(x-1/x)+½x²(2x/(x²-1)-1/x)+ln(2),

f'(x)=xln(x-1/x)+x³/(x²-1)-x/2+ln(2).

Newton's iterative Method:

x_n+1=x_n-f(x_n)/f'(x_n). We need to choose x₀ to start the process. We must avoid x₀=0 and x₀-1/x₀≤0 which would cause anomalies in the expressions. Choose x₀=1.5.

x₁=1.00913, x₂=1.02155..., ..., x=1.03681291616 after a few more iterations.

However, this is not the only solution, so we need to set x₀ to a different value nearer the other solution. A rough sketch shows that x₀=-0.75 is a promising start value. After a few iterations, x=-0.75871567976.

So there are two solutions as calculated above.