Monic to Manic!

$(x^{2}-a^{2})^{50}=(x-a)^{50}(x+a)^{50}$ . Thus, of the $100$ roots of the expression, $50$ are $+a$ , and $50$ are $-a$ . Now, the coefficient of $x^{99}$ is, by Vieta's Formula, the negative of the sum of all the roots, since the polynomial is monic. The sum of the roots is $0$ . Now, since the power of $x$ is $2$ , every coefficient of a power of $x$ is non-zero, for even powers of $x$ . Thus, all coefficient of an odd power of $x$ is $0$ . Thus, the coefficient of $x^{49}=0$ . Now, the constant is just $(a^{2})^{50}=a^{100}$ . Thus, $b=0+0+a^{100}$ . Now, $\log_ab=\log_aa^{100}=100\log_aa=\boxed{100}$