How to solve transcendental equation Is this a transcendental equation?
$$100 \cdot n^2 = 2^n$$
Can you provide a step-by-step solution, please?
 A: This a transcendental equation and the solution cannot express in terms of elementary functions.
But, for your curiosity, any equation which can write as $$A+Bx+C\log(D+Ex)=0$$ has solutions in terms of Lambert function $W(z)$ which such that $z=W(z)\,e^{W(z)}$. 
In the case you gave, there are three solutions which write $$n_1=-\frac{2 W\left(-\frac{\log (2)}{20}\right)}{\log (2)}\approx 0.103658$$ $$n_2=-\frac{2 W\left(\frac{\log (2)}{20}\right)}{\log (2)}\approx -0.096704$$ $$n_3=-\frac{2 W_{-1}\left(-\frac{\log (2)}{20}\right)}{\log (2)}\approx 14.3247$$
If you cannot use Lambert function, only numerical methods (such as Newton) will give he numerical solution. But, a "reasonable" guess of the solution is required. Then, plot the function $$f(n)=100\,n^2-2^n$$ or much better (if you are only concerned by positive values) $$g(n)=\log(100)+2\log(n)-n\log(2)$$ locate an approximate value and start Newton iterations.
The derivative $$g'(x)=\frac{2}{n}-\log (2)$$ cancels for $n_*=\frac{2}{\log (2)}$ and $g(n_*)=-2+\log (100)+2 \log \left(\frac{2}{\log (2)}\right)\approx 4.72449$. Since the second derivative is always negative, this is a maximum and there are two roots. Plotting the function and being very lazy, let us not a solution around $10$; so, we shall start iterations using $n_0=10$. Newton method will update the guess according to $$n_{k+1}=n_k-\frac{g(n_k)}{g'(n_k)}$$ which in your case write, after simplifications,  $$n_{k+1}=\frac{2 n_k (\log (n_k)-1+\log (10))}{n_k \log (2)-2}$$ and the iterates will be $$n_1=14.6211$$ $$n_2=14.3255$$ $$n_3=14.3247$$ which is the solution for six significant figures.
A: This transcendental equation can be solved with the Lambert W function. First an easy transform
$$100 n^2 = 2^n\Longrightarrow 10 n = \pm\sqrt{2^n}=\pm 2^{n/2}$$
Substitute $t=n/2$ and solve $20 t = 2^t$ using the Example 1
and get 
$$t_1 = -\frac{W_0(-\frac{\ln 2}{20})}{\ln 2}$$
$$t_2 = -\frac{W_{-1}(-\frac{\ln 2}{20})}{\ln 2}$$
and from $20 t = -2^t$
$$t_3 = -\frac{W_0(\frac{\ln 2}{20})}{\ln 2}$$
The solutions for $n$ are $n_i=2t_i$ with the appoximate values
$$n_1= 0.1036578164,\quad n_2=14.324727837,\quad n_3=-0.0967040343267$$
