>>12474410This guy has the right idea: Apply the log2 function to both sides when you can and rearrange based on the identity log2(a^b) = b*log2(a) (a & b > 0) and drop out the slowest growing parts of the resulting equations.
a left = log2(log2(n)*log2(n)
the log2(n) dominates this at high n so we can approximate it has just log2(n)
a right = log2(n)^2 * log2(2)
we can approximate this as log2(n)^2 for large n.
Clearly log2(n) is going to grow slower than log2(n)^2 so right is the answer
b left = 1.0001^0.001n
No need to log here, we can appoximate this has 1.0001^n for large n
b right = n^1001 + n^log2(n)
we can approximate this as n^log2(n)
1.0001^n grows faster than n^log2(n) so left is the answer.
c left = log2(n)*log2(log2(log2(n))
c right = log2(log2(log2(n)))*log2(n)
these are the same function so the answer is equal.
d left = 100000*log2(log2(n))
we can approximate this has log2(log2(n))
d right = 0.5*log2(n) (since sqrt(n) = n^0.5)
we can approximate this has log2(n) and clearly this grows faster than log2(log2(n)) so the answer is the right.
