Which of the numbers is larger: $7^{94}$ or $9^{91} $?

Solution 1:

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 $343$.

Solution 2:

$$7^{94} = 7^{10} 49 ^{42} < 7^{10} 54 ^{42} = 7^{10} 8^{14} 9^{63} < 9^{10} 9^{14} 9^{63} = 9^{87} < 9^{91} $$

Solution 3:

$Log(9^{91})=91\cdot Log(9)=86.836068359$

$Log(7^{94})=94\cdot Log(7)=79.4392157613$.

Hence $ 9^{91}$ is bigger.

Solution 4:

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}

Solution 5:

$9^{91} \div 7^{94} = (\frac97)^{94} \div 9^3 > (1+\frac27)^{7 \times 13} \div 3^6 > (1+2)^{13} \div 3^6 = 3^7$ which is way bigger than $1$.