The math help thread

Graphed 2^(n^2)-2^n-5=0 and found the positive zero with the graphing calculator's zero thing. 1.75. o.0
Seems a bit off from 2.10.

Are you sure you can take a log of each term on one side like that?
ln(5+7)-(ln(5)+ln(7))=-1.07. That doesn't seem equal to me.

This is correct, you cannot apply logarithms like I did. My, my, I seem to go old.

let a=2^n
thus the equation a^2-a=5
solve a (do it urself i'm too lazy)
then n= log a/ log 2

You cannot do it that way. If a = 2^n, then 2^(n^2) = a^n. (By the way: Are you 100% sure it's 2^(n^2) and not (2^n)^2? )

[spoiler=Scrabbles, most won't likely help any]
I'll see what I'll be able to do for tomorrow, as it's gettin late over here[/spoiler]

Noah