For this example simple fixed-point iteration works. Reformulate the problem, such that $x=F(x)$. One(!) example of doing so would be \begin{align} x = \frac{1}{(4\log_{10}(Re\cdot \sqrt{x})-0.4)^2}=F(x) \end{align} Fixed-point iteration works like this: $x^{k+1}=F(x^k)$. That means you simply plug in the result of $F(x^k)$ once more into $F$. Now you have to prove that this does converge, see i.e. Banach fixed-pint theorem.
Another method to use would be Newton's Method, but it turns out that the fixed-point iteration just works fine. With the Matlab code below, you reach a fixed-point in 5 iterations.
function fixedPointIteration x = 1; tol=1e-5; while abs(x-f(x))>tol i=i+1 x=f(x); end sprintf('Result of fixed-point iteration is x=%s',x) end function y=f(x) Re = 1e4; y =1/(4*log10(Re*sqrt(x)-0.4))^2; end
Which gives $x=0.0072742$. It should be easy to adapt the code into other languages. Hope this helps.