How to solve this equation for $n$: $n \le 2^n a$?
Note: $a$ is a positive constant.
My Method: $\\ n \leqslant 2^n a\\ \lg {n} \leqslant \lg{(2^n a)}\\ \lg{n} \leqslant \lg{a} + \lg{2^{n}}\\ \lg{n} - \lg{2^{n}} \leqslant \lg{a}\\ \lg{(\frac n {2^n·})} \leqslant \lg{a}\\ \frac n {2^n} \leqslant a$
I just reached the question statement. Unable to figure out how to solve this.
The origin of this question: I was trying to find the Big-$O$ complexity of a relation which demands to solve an inequality. The original value of $a$ is $2^{10}$. I want to know how to solve this equation to calculate the range of values of $n$ for which the above inequality is true.
Let $f(x)=x$ and $g(x)=a.2^x$
Its obvious that for all negative $x$, $g(x)$ lies above $f(x)$. (If $a>0$) Let's find where these two curves touch each other.
For touching case, let the curves meet at $(x_1,y_1)$. If they touch each other, then slope of tangent at that point is same for both curves. That means, $$1=a\cdot 2^{x_1}\log 2$$ and $$2^{x_1}=\frac{1}{a\log 2}$$
$x_1$ satisfies $f(x)=g(x)$ as well. So, $$x_1=a.2^{x_1}$$ $$x_1=\frac{1}{\log 2}$$
Plug this back in the first equation to get $$a=\frac{1}{2^{\frac{1}{\log(2)}}\cdot \log(2)}$$ Hence, the two curves touch each other for the above value of $a$.
This gives you three cases.
For $a$ equal to the above value, the curves only touch each other.
For $a$ greater than the calculated value, there will be no solution
For $a$ lesser than the calculated value, then two solutions exist.