solve equation with logarithm base 2

Question

Solve in IR: I tried but traditionally couldn t be solved.

$\sqrt{x-\sqrt{x-1}} + \sqrt{x+\sqrt{x-1}} +log_2(x-1)=0$

Answer 1

As Gerry Myerson commented, there is no closed form expression of the zero of function $$f(x)=\sqrt{x-\sqrt{x-1}} + \sqrt{x+\sqrt{x-1}} +\log_2(x-1)$$ By inspection, we know that the solution is between $x=1$ and $x=2$ since $f(2)=1+\sqrt{3}$.

We can have a quite good approximation of the solution using the Taylor expansion at $x=1$ $$\sqrt{x-\sqrt{x-1}} + \sqrt{x+\sqrt{x-1}}=2+\frac{3 (x-1)}{4}+O\left((x-1)^{2}\right)$$ which makes, at least locally,

$$f(x) \approx \frac{3 (x-1)}{4}+\frac{\log (x-1)}{\log (2)}+2$$ the solution of which being given in terms of Lambert function. $$x=1+\frac{4 }{3 \log (2)}W\left(\frac{3 \log (2)}{16}\right)$$ Since the argument is quite small, you can evaluate $$W(z)=z-z^2+\frac{3 z^3}{2}-\frac{8 z^4}{3}+\frac{125 z^5}{24}+O\left(z^6\right)$$ which is given in the linked page. Using it, the approximate solution is then $\approx 1.22267$ while the "exact" solution would be $\approx 1.22250$.

If you need more accuracy, just use Newton method with, for example, $x_0=1.5$ and get the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.5000000000000000000 \\ 1 & 1.1203284150829764246 \\ 2 & 1.1959735634030846971 \\ 3 & 1.2210074897823484868 \\ 4 & 1.2224983881275373905 \\ 5 & 1.2225029048928075835 \\ 6 & 1.2225029049338946555 \end{array} \right)$$ which is the solution for twenty significant figures.

Edit

An interesting (at least to me) generalization would be to find the zero of $$g(x)=\sqrt{x-\sqrt{x-1}} + \sqrt{x+\sqrt{x-1}} +\log_a(x-1)$$ Using the same procedure, the approximate solution would be given by $$x=1+\frac{4 }{3 \log (a)}W\left(\frac{3 \log (a)}{4a^2}\right)\tag 1$$ which can be approximated by the simple $$x=1+\frac 1{a^2}-\frac{3 \log (a)}{4 a^4}+\frac{27 \log ^2(a)}{32 a^6}+\cdots\tag 2$$

Looking at a few values of $a$, here are the approximate and exact results $$\left( \begin{array}{cccc} a & \text{from } (1) & \text{from } (2)& \text{exact} \\ 2 & 1.22267 & 1.22384 & 1.22250 \\ 3 & 1.10214 & 1.10234 & 1.10210 \\ 4 & 1.05879 & 1.05883 & 1.05878 \\ 5 & 1.03820 & 1.03821 & 1.03819\\ 6 & 1.02680 & 1.02680 & 1.02679 \\ 7 & 1.01983 & 1.01983 & 1.01983 \\ 8 & 1.01526 & 1.01526 & 1.01526 \\ 9 & 1.01210 & 1.01210 & 1.01210 \\ 10 & 1.00983 & 1.00983 & 1.00983 \end{array} \right)$$