Which of the numbers is larger: $7^{94}$ or $9^{91} $? In this problem, I guess b is larger, but not know how to prove it without going to lengthy calculations. It is highly appreciated if anyone can give me a help.

Which number is larger
$$\begin{align}
  &\textrm{(a)}\quad 7^{94}   
  &\quad\textrm{(b)}\quad 9^{91} 
\end{align}$$  


I voted for André's answer, but here's another approach, using a different bit of maths.
Note that $7^{94} = 7^3 \times 7^{91}$. $9^{91} = (7 \times \frac{9}{7})^{91}$, where $\alpha = \frac{9}{7} = 1 + \frac{2}{7} > 1$. So
$$
\frac{7^{94}}{9^{91}} = \frac{7^3}{\alpha^{91}}.
$$
What do we make of $\frac{7^3}{\alpha^{91}}$? Well, $7^3 = 49 \times 7 = 343$. Using the binomial theorem, and observing that positive ratios always diminish when the numerators (resp. denominators) are decreased (resp. increased),
\begin{align*}
\alpha^{91} &= \left(1 + \frac{2}{7}\right)^{91} \\
 &> 1 + \frac{91}{1!} \times \frac{2}{7} + \frac{91 \times 90}{2!} \times \frac{2^2}{7^2} \\
 &\quad= 1 + \frac{182}{7} + \frac{8190 \times 4}{98} \\
 &\quad> 1 + 25 + \frac{4 \times 80 \times 100}{100} \\
 &\qquad= 1 + 25 + 320 \\
 &\qquad= 346 \\
 &\qquad> 343.
\end{align*}
So $\alpha^{91} > 7^3$ and thus $9^{91} > 7^{94}$.
