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

 A: $Log(9^{91})=91\cdot Log(9)=86.836068359$
$Log(7^{94})=94\cdot Log(7)=79.4392157613$.
Hence $ 9^{91}$ is bigger. 
A: André already nailed it, but here's another way. The following inequalities are equivalent:\begin{align}7^{94} &< (7+2)^{94-3} \\ 9^3&<(1+2/7)^{94} \\ 3\log3&<47\log(1+2/7),\end{align} and by the Maclaurin expansion of $\log(1+x)$, the latter follows from \begin{align}3\log3&<94\left(\frac17-\frac1{49}\right) \\ \log3<3&<2\cdot\frac{94}{49}.\end{align}
A: The first is $7^{91}\times 343$. The second is $7^{91}\times(9/7)^{91}$. Since $\frac{9^3}{7^3}\gt 2$, it follows that $(9/7)^{91}$ is much much bigger than $343$.
A: $9^{91} \div 7^{94} = (\frac97)^{94} \div 9^3 > (1+\frac27)^{7 \times 13} \div 3^6 > (1+2)^{13} \div 3^6 = 3^7$ which is way bigger than $1$.
A: First notice that $3^9 = 19683 > 16807 = 7^5$ (this can be calculated manually).
Thus $9^9 = (3^2)^9 = 3^{18} = (3^9)^2 > (7^5)^2 = 7^{10}$.
It follows that $9^{91} > 9^{90} = (9^9)^{10} > (7^{10})^{10} = 7^{100} > 7^{94}$.
A: $$7^{94} = 7^{10} 49 ^{42} < 7^{10} 54 ^{42} = 7^{10} 8^{14} 9^{63} < 9^{10} 9^{14} 9^{63} = 9^{87} < 9^{91} $$
A: 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 $9^{91} > 7^{94}$.
A: Here are a couple of equivalent statements:
\begin{eqnarray*}
7^{94}&<&9^{91}\\
7^3&<&\left(\frac97\right)^{91}\\
343&<&\left(1+\frac27\right)^{91}
\end{eqnarray*}
We can expand the latter expression using the binomial theorem to get
$$\left(1+\frac27\right)^{91}>1+\binom{91}{1}\times\frac27+\binom{91}{2}\times\left(\frac27\right)^2=1+26+\frac{90\times26}{7}>343.$$
