Multiplying exponents, solving for n

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:

$$100n^2<2^n$$

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!

Tags:

Algorithms

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