TITLE: Multiplying exponents, solving for n
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: One of the answers says that problem can be stated as an inequality:
$$100n^2<2^n$$
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!