# solve equation with logarithm base 2

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

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

• It's not a linear equation. I'm guessing it can't be solved in closed terms, but only solved numerically. Apr 25, 2018 at 2:12
• By the way, write $\$$\log\$$ to generate$\log$as opposed to$log$. Apr 25, 2018 at 2:55 • Why? It denotes a base other than 10, only Log to denote base 10. – Papa Apr 25, 2018 at 4:38 ## 1 Answer 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 $$\left( \begin{array}{cccc} a & \text{from } (1) & \text{from } (2)& \text{exact} \\ 2 & 1.22267 & 1.22384 & 1.22250 \\ 3 & 1.10214 & 1.10234 & 1.10210 \\ 4 & 1.05879 & 1.05883 & 1.05878 \\ 5 & 1.03820 & 1.03821 & 1.03819\\ 6 & 1.02680 & 1.02680 & 1.02679 \\ 7 & 1.01983 & 1.01983 & 1.01983 \\ 8 & 1.01526 & 1.01526 & 1.01526 \\ 9 & 1.01210 & 1.01210 & 1.01210 \\ 10 & 1.00983 & 1.00983 & 1.00983 \end{array} \right)$$

• Neatly done! Thanks
– Papa
Apr 25, 2018 at 4:38
• I had someone done a great job solving the equation by removing the square roots and leaving an absolute value of both and creating two intervals, by which you only get an answer in one interval, x= 5/4, why there would be a need to take those square roots?
– Papa
May 13, 2018 at 17:38
• @J.Moh. Could you give more details ? As far as I can see, there is only one real root to the equation. May 14, 2018 at 3:05
• @ Claude Leibovici. math.stackexchange.com/questions/2915745/…
– Papa
Sep 13, 2018 at 15:31