Is there a numerical answer for $\sqrt[n]{2^{3.6\times 10^{13}}}$

Question

I am trying to see if there is an actual numerical solution for $n$ in the equation $n = \sqrt[n]{2^{3.6\times 10^{13}}}$. According to the solutions manual for an algorithm book and chegg, $n = 9 \times 10^{11}$. How did they arrive at this solution for $n$? It is confusing knowing that $n$ has to be evaluated to something to get an approximation, but how are they approximating it?

The question originally asked was that a computer can perform $10^{10}$ operations per second. Now it is asking what is the largest input size $n$ for which you could obtain a result in 1 hour. $n\log n$ is one of the input sizes. This was what I did

$ n = 60 \times 60 \times 10^{10} = 3.6^{13}$, how much it can perform in a hour. And,

$n\log n = 3.6^{13}$

$2^{n\log n} = 2^{3.6^{13}}$

$n^{n} = 2^{3.6^{13}}$

$n = \sqrt[n]{2^{3.6\times 10^{13}}}$

Answer 1

Hint : Logarithming the equation gives $$\ln(n)=\frac{3.6\cdot 10^{13}\cdot \ln(2)}{n}$$ This equation is easier to analyze.

Answer 2

I would convert $n = \sqrt[n]{2^{3.6\times 10^{13}}}$ into $n^n = 2^{3.6\times 10^{13}}$ first and take $\ln$ to get $$f(n):=n\ln n - 3.6\times10 ^{13} (\ln 2)=0.$$

Now I would just use a calculator plugging in various large values of $n$ till the sign of $f(n)$ switches from negative to positive and just keep narrowing the interval down to your heart's content/desired accuracy.

Admittedly not the greatest solution but it seems simple enough.

Answer 3

With $t = 3.6 \times 10^{13}$, your equation is $n = 2^{t/n}$. This can be solved with the Lambert W function:

$$ n = \frac{t \ln 2}{W(t \ln 2)} $$

The numerical value is approximately $9.063164830 \times 10^{11}$.

Answer 4

As Ravi wrote, consider $$f(n)=n\log(n)-36\times 10^{12}\log(2)$$ and you look for the zero of it.

Remembering that $n\log(n)>n$, this gives an upper bound equal to $36\times 10^{12}\log(2)$. Using this as $n_0$, start using Newton method. The successive iterates will be $$\left( \begin{array}{cc} k & x_k \\ 0 & 2.4953298500158031139\times 10^{13} \\ 1 & 1.5670231876732479624\times 10^{12} \\ 2 & 9.1197181285758111379\times 10^{11} \\ 3 & 9.0631709855044190836\times 10^{11} \\ 4 & 9.0631648285318432104\times 10^{11} \\ 5 & 9.0631648285317699141\times 10^{11} \end{array} \right)$$ which is the solution for twenty significant figures.

More generally, if you need to solve $$g(n)=n\log(n)-a$$ where $a$ is large, you can get good approximations using approximation of Lambert function. Since $n=\frac{a}{W(a)}$ (as Robert Israel already answered), use $$W(a)=L_1-L_2+\frac {L_2}{L_1}+\frac {L_2(L_2-2)}{2L_1^2}+\frac{L_2(2L_2^2-9L_2+6)}{6L_1^3}+\cdots$$ where $L_1=\log(a)$ and $L_2=\log(L_1)$.

Adding the successive terms one at the time, we should have for $a=36\times 10^{12}\log(2)$ $$\left( \begin{array}{ccc} k & W(a)\approx & n \approx \\ 1 & 30.8480 & 8.08911\times 10^{11} \\ 2 & 27.4190 & 9.10075\times 10^{11} \\ 3 & 27.5301 & 9.06400\times 10^{11} \\ 4 & 27.5327 & 9.06315\times 10^{11} \\ 5 & 27.5327 & 9.06316\times 10^{11} \end{array} \right)$$