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
$$\begin{align} &\textrm{(a)}\quad 7^{94} &\quad\textrm{(b)}\quad 9^{91} \end{align}$$
On
$Log(9^{91})=91\cdot Log(9)=86.836068359$
$Log(7^{94})=94\cdot Log(7)=79.4392157613$.
Hence $ 9^{91}$ is bigger.
On
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}
On
$$7^{94} = 7^{10} 49 ^{42} < 7^{10} 54 ^{42} = 7^{10} 8^{14} 9^{63} < 9^{10} 9^{14} 9^{63} = 9^{87} < 9^{91} $$
On
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 $9^{91} > 9^{90} = (9^9)^{10} > (7^{10})^{10} = 7^{100} > 7^{94}$.
On
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 $9^{91} > 7^{94}$.
On
$9^{91} \div 7^{94} = (\frac97)^{94} \div 9^3 > (1+\frac27)^{7 \times 13} \div 3^6 > (1+2)^{13} \div 3^6 = 3^7$ which is way bigger than $1$.
On
$$\begin{align} \left(9\over7\right)^3={729\over343}\gt2 &\implies\left(9\over7\right)^{15}\gt2^5\gt7\\ &\implies9^{15}\gt7^{16}\\ &\implies9^{90}\gt7^{96}\\ &\implies9^{91}\gt7^{94} \end{align}$$
Another proof, using the general inequality $\ln(1-x)\lt-x$ for $0\lt x\lt1$ and the numerical inequality $7\lt2^3\lt e^3$:
$$\ln(7^{94}/9^{91})=3\ln7+91\ln\left(1-{2\over9}\right)\lt3\cdot3-90\cdot{2\over9}=9-20\lt0$$
A third proof, presented in easily checkable, but almost completely unmotivated form:
$$\begin{align} 2^{47}7^{94} &=98^{47}\\ &\lt100^{47}\\ &=10^3\cdot10^3\cdot10^{88}\\ &\lt2^{10}\cdot2^{10}\cdot10000^{22}\\ &\lt20000^{22}\\ &\lt160^{44}\\ &\lt36(162)^{45}\\ &=2^{47}9^{91} \end{align}$$
And one more proof, this one based on the fact that $2^{10}=1024\gt1000=10^3$, which implies $\log2\gt0.3$ (where "log" here means log base $10$):
$$94\log7=47\log49\lt47(\log100-\log2)\lt47\cdot1.7=79.9$$ whereas
$$91\log9\gt91\log8=273\log2\gt273\cdot0.3=81.9$$
Full disclosure: I used a calculator for $47\cdot1.7=79.9$. But everything else I did by hand.
Added 5/25/15: At another question, proofs are given of the inequality $7^{19}\lt9^{17}$. (See in particular joriki's answer there.) It follows that
$$7^{94}\lt7^{95}\lt9^{85}\lt9^{91}$$
On
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 $$\left(1+\frac27\right)^{91}>1+\binom{91}{1}\times\frac27+\binom{91}{2}\times\left(\frac27\right)^2=1+26+\frac{90\times26}{7}>343.$$
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 $343$.