Richard Heathfield wrote:

> Dik T. Winter said:

>> (E-Mail Removed)lid writes:

>>

>>> It may help the OP to recognise that the question, as asked (i.e.

>>> in the absence of big-O notation, which IMHO would in any case

>>> make the question meaningless), is looking for the integer value

>>> of n that is just higher than the real value that is the solution

>>> to the following equation:

>>>

>>> 2 to the power n = 100 * n * n

>>>

>>> This is a simple mathematical exercise.

>>

>> Finding the real value is not exactly a simple mathematical

>> exercise.

>

> It isn't? Using a simple iterative technique, I found the answer

> in nothing flat (actually about 3 milliseconds), getting agreement

> in 2^n and 100n^2 to ten decimal places. Given that we only *need*

> it to one decimal place, I'd have thought that was an adequate

> solution.
Taking log2() function of both sides, we have:

n = log2(100) + 2 * log2(n)

which makes it fairly easy to divide integers into two classes,

i.e. those where "n < log2(100) + 2*log2(n)" and those where "n >

log2(100) + 2*log2(n)". Note that the equality condition cannot

occur since 2 does not contain all the factors of 100. We can go

close to exactness by considering 100 as 2*2*5*5, so that log2(100)

= 2 + 2* log2(5), and that result is obviously greater than 6 and

less than 7.

All this should be obvious to any child of 10 who has passed the

kindergarten class on logarithms.

--

Chuck F (cbfalconer at maineline dot net)

Available for consulting/temporary embedded and systems.

<http://cbfalconer.home.att.net>

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