The math help thread

Oh but you also can't separate log(1+-sqrt(21)) into log(1) +- log(sqrt(21)). heh.

And since log of a negative number is complex, the only real solution would be n = log2(1 + sqrt(21)) = ln(1 + sqrt(21))/ln(2) =~ 1.481

That certainly is correct. I was thinking about it at university when I had my math-class. I don't understand how I manage to do the same math-error twice, but oh well.

[IMGspoiler]http://i36.tinypic.com/33ljl0k.jpg[/IMGspoiler]

Anywho, for those who are interested in the

answer: Do the following:

Set


First off, prove that there exist a real answer (A prof. and I discussed the topic, and we got off to imaginary numbers. Yehaw.):
As

is true for all real a, the function is continuous. (Or use Cauchy, if you're that unsure that this is a continuous function)
Set n=1, then f(n) = 0. If we set n = 2, then


Therefore, there is a root 1<n<2 such that f(n) = 5.

To find the exact answer, you can turn it to a series expansion or repeat Newton's method infinitely many times (Well...).

Noah