From Rosen's Discrete Mathematics and Its Applications, 3ed, chapter 8.3 p. 530:
From the underlined info, this is what I believe they get:
$\ n^{log_b (a)}, n = b^{k} \Rightarrow n^{log_b (a)} = (b^{k})^{log_b (a)} = a^{k}$
Can someone explain to me the part where they end up with$\ a^{k}$ ? This is a bit convoluted for me. Thank you!
If we assume that $n=b^k$, for $b \in \mathbb{Z}$, $b > 1$, then we directly obtain that $$ n^{\log _b(a)} = (b^k)^{\log _b (a)}. $$ Now, since $\log _b (a^k) = k \cdot \log _b (a)$ (a property of logarithms), we have $$ (b^k)^{\log _b (a)} = b^{\log _b (a^k)}. $$ Finally, since $\log _b (x)$ is the inverse function of $b^x$, we have that $$ b^{\log _b (a^k)} = a^k, $$ provided that $a^k$ is positive (which it is).