Getting n alone in n=x and nlgn=x

[Dafydd]

Homework Statement

I need to know the greatest n for which n! is greater or equal to x, and the greatest n for which nlgn is greater or equal to x.

Homework Equations

Can't think of any. Sorry!

The Attempt at a Solution

For a "normal" function, like, say, lgn or n^2, it would be as easy as inverting the function and getting n on its own, like this:

lgn = x
n = 2^x (where lg is of base 2)

or

n^2 = x
n = sqrt(x).

However, for n!=x and nlgn=x I can't come up with a way to do this. Help!

 

[mighty2000]

I would like to provide some guidance on how to approach this problem. First, let's define n! and nlgn to make sure we are on the same page. n! (n factorial) is the product of all positive integers from 1 to n. For example, 5! = 1x2x3x4x5 = 120. On the other hand, nlgn is a logarithmic function where the base of the logarithm is n. So, if we write it as nlgn = x, it means nlogn = x.

Now, let's look at the problem at hand. We need to find the greatest n for which n! is greater or equal to x, and the greatest n for which nlgn is greater or equal to x. In other words, we need to find the value of n when the value of n! or nlgn reaches or exceeds x.

For n!, we can use the Stirling's approximation, which states that for large values of n, n! is approximately equal to √(2πn)(n/e)^n. This approximation can give us a good estimate for the value of n when n! equals x.

For nlgn, we can use the inverse of the logarithm function to find n. The inverse of logarithm is the exponential function, so we can write it as n = e^x. This means that when nlgn equals x, n is equal to e^x.

I hope this helps you in finding the solutions to your problem. Remember, as a scientist, it is important to understand the underlying concepts and use appropriate approximations or equations to solve a problem.