The number of ordered pairs $(a, b)$ of integers which satisfy the equation $a^{b}=64$ is
(A) 3
(B) 5
(C) 8
(D) 6
(E) 7
We must find all ordered pairs $(a, b)$ of integers which satisfy $a^{b}=64=2^{6}$. We can first say that we know that $b$ must be positive, since $a^{b}$ is bigger than 1 . However, it is possible that $a$ is negative.

If we want both $a$ and $b$ to be positive, the only ways to write 64 are $64=64^{1}=8^{2}=4^{3}=2^{6}$ (since $a$ must be a power of 2). If $b$ is even, then it possible for $a$ to be negative, ie. $64=(-8)^{2}=(-2)^{6}$. So there are 6 ordered pairs $(a, b)$ that satisfy the equation.