When solving for n in this equation I get stuck.

Question: What is the smallest value of n such that an algorithm with running time of $\ 100n^2 $ runs faster than an algorithm whose running time is $\ 2^n $ on the same machine?

Straight out of CLRS chapter 1. Class starts in 2 months wanted to get a head start.

My approach:

$\ 100n^2 = 2^n $

$\ \sqrt(100n^2) = \sqrt(2^n) $

$\ 10n = (2^n)^{1/2} $

$\ 10n = (2^{n/2}) $

Is this last step correct? I know I add exponents when multiplying but this is raising an exponent to an exponent so I should multiply. I'm still unsure how to bring the n down out of the exponent on the two so I can solve for it.

  • $\begingroup$ $\sqrt{(100n^2)} \neq 10n^{2}$. But what you've done with the exponent at the end is right. $\endgroup$ Aug 13, 2012 at 17:58
  • $\begingroup$ ahh good catch thanks. $\endgroup$ Aug 13, 2012 at 17:59
  • $\begingroup$ The fact that $10n=2^{n/2}$ may help a little. But there is no way to solve the resulting equation by "formula." (I am lying slightly.) You will need to fool around a bit with numbers to get an answer. $\endgroup$ Aug 13, 2012 at 18:02
  • $\begingroup$ I can't divide by 2 either since the expression on the right contains a variable in the exponent. Any hints on how I can express $\ 2^n $ in a way that makes it more conducive to be solved by 10n? $\endgroup$ Aug 13, 2012 at 18:05
  • $\begingroup$ There's no easy route that I am aware of. Just compute the numbers. $\endgroup$
    – copper.hat
    Aug 13, 2012 at 18:06

5 Answers 5


Sometimes in cases like this, it's useful to simply try a few and see what happens:

Trying $n=20$ gives $$10n=200$$ and $$2^{n/2}=2^{10}=1024.$$

Since the rate of increase of $n\mapsto 10n$ is constant, while $n\mapsto 2^{n/2}$ grows more and more quickly for larger $n,$ it follows that $10n<2^{n/2}$ whenever $n\ge 20,$ so we're looking for some $n<20.$

Trying $n=14$ gives us $$10n=140$$ and $$2^{n/2}=128,$$ so by similar reasoning, we need some $n\ge 14$.

From there, the solution's fairly quickly found to be $n=15$.

Note: $n=15$ is not the solution to the equation $100n^2=2^n$, but it is the least $n$ for which $100n^2<2^n$.


One of the answers says that problem can be stated as an inequality:


Take $\log_{2}$ of both sides:

$$\log_2{(100n^2)} < n$$

By logarithm rules:

$$\log_2{(100)} + \log_2{(n^2)} < n$$

You may remove the squared term (I'm not sure if this matters):

$$\log_2{(100)} + 2\log_2{(n)} < n$$

Rearrange by subtracting $ 2\log_2{(n)}$ from both sides:

$$\log_2{(100)} < n- 2\log_2{(n)}$$

Square both sides:

$$(\log_2{(100)})^2 < (n- 2\log_2{(n)})^2$$

Expanding out the brackets, this may be expressed as a quadratic:

$$ n^2 - 4\log_2{(n)}n + 4(\log_2{(n)})^2 - (\log_2{(100)})^2 > 0$$

Now we must solve for $n$.

Wolfram Alpha

one of the solutions is $ n > 14.3247$ which is what one of the other answers shows in the linear plot.

This might be a very convoluted way to go about it, _'m not sure. But the other answer involved plotting and looking to narrow down. That might be found with newton iteration algorithm as well. But this quadratic way I think is a direct way to get the answer. The other root is less than 1 so I guess it can be ignored.

I'm very happy to be shown if there's a simpler way!


This is correct. You can then start guessing to find it by hand. Clearly for $n=20$ left hand side is much smaller than right hand side. For $n=16$ similarly, $n=15$ seems okay, since $2^7=128$ and $\sqrt 2>1.4$, and $n=14$ is too small for similar reasons.


The equation $10 n = 2^{n/2}$ can't be solved in terms of elementary functions: you have to use the Lambert W function. The two real solutions are $-2 LambertW(-\ln(2)/20)/\ln(2)$ and $-2 LambertW(-1,-\ln(2)/20)/\ln(2)$. But for what you want to do, this is not very relevant. "Guess and check" is the way to go.


I used Wolfram to narrow down on the approximate solution. I started with:

plot 2^x - 100 *x^2 , x=1 to 100

and looked at the graph to narrow the value of n down to:

plot 2^x - 100 *x^2 , x=14.3245 to 14.325


after a couple steps.

This was after trying to reduce it to the simplest terms though my knowledge doesn't make it to Lambert functions.


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