how to solve $2169-2^n-n^2=0$ I need to solve this equation: $2169-2^n-n^2=0$
So I have tried to guess a solution for maybe checking by derivative that it's the only one.
I didn't succeeded.
Thanks.
 A: If you sketch the graphs of the exponential curve $y=2^x$ and the downward-pointing parabola $y=2169-x^2$ (or just look at mvw's answer, which was posted simultaneously with this one), you will see they have exactly two points of intersection, one with $x\gt0$ and one with $x\lt0$.  As luck would have it, the positive crossing occurs at $x=11$ precisely.  Given the fact that the exponential curve tends fairly rapidly to $0$ for large negative values of $x$, it's clear that the negative crossing occurs at around $x=-\sqrt{2169}\approx-46.57$.
A: The equation
$$
2169 - n^2 = 2^n
$$
can be viewed as intersection of the graphs of $f(x) = 2169 - x^2$ (green parabola) and $g(x) = 2^x$ (red exponential function).

You can use this for a graphical solution, but for precise values some numerical method seems necessary.
A: $n=11$ because
$$121-n^2=2^{11}(2^{n-11}-1)$$
$$(11-n)(11+n)=2^{11}(2^{n-11}-1)$$
if $n> 11$ then $(11-n)(11+n)<0$ and $\,2^{11}(2^{n-11}-1)>0$.
if $-11<n< 11$ then $(11-n)(11+n)>0$ and $\,2^{11}(2^{n-11}-1)<0$.
if $n\le-47$ then $2169-n^2<0$  and $\,2^n>0$
if $-46\le n\le-11$ then $2169-n^2>2^n$
$-\sqrt{2169}$ is negative solution of $f(n)=20169-n^2$ then $y_1=2^n$ and $y_2=2169-n^2$ cross toghether in $(-47,-46)$
A: Equations like $$2^x+x^2=2169$$ cannot in general be solved exactly  by elementary ways. Searching for approximations we find out by chance $x=11$ as an exact solution and, because of the graphic of the function $f(x)=2^x +x^2$ we know that for all $y\ge 1$ (actually for $y\ge y_m$ where $y_m$ is less that $1$ and it is the minimun of the function $f$) there are two points $x_1,x_2$ such that $f(x_1)=f(x_2)=y$.
The corresponding other point to the exact solution $x=11$ is such that it is between $-46$ y $-47$ because $f(-46)=2116 +\epsilon$ and $f(-47)=2209+\epsilon'$ where $\epsilon$ and $\epsilon'$ are very small.(This problem of determine the exact value of this point is the kind of problem we would have had if we do not have the exact solution $x= 11$). 
A: Consider the function $$f(x)=2^x+x^2-a$$ $$f'(x)=2 x+2^x \log (2)$$ $$f''(x)=2^x \log ^2(2)+2$$ If there is no analytical solution for the roots of $f(x)=0$, there is one analytical solution for $f'(x)=0$. Using Lambert function, the first derivative cancels for $$x_*=-\frac{W\left(\frac{\log ^2(2)}{2}\right)}{\log (2)}\approx -0.284538$$ Using the above result $$f(x_*)\approx 0.901966 - a$$ and $f''(x)$ is always positive. So, if $f(x)<0$, there are two roots for equation $f(x)=0$ (one positive and one negative).
If $a$ is large, the positive root is approximately given by $2^x=a$ that is to say $x=\frac{\log (a)}{\log (2)}$ and the negative root is approximately given by $x^2=a$ that is to say $x=-\sqrt a$.
Applied to the case where $a=2169$, this gives as approximations $x_1\approx 11.0828$ and $x_2\approx -46.5725$ while the exact solutions are $11$ and $\approx -46.5725$.
Using these simple estimates, you could start Newton method to polish the roots.
