How can I prove that $7^{31}$ is bigger than $8^{29}$?

I tried to write exponents as multiplication, $2\cdot 15 + 1$, and $2\cdot 14+1$, then to write this inequality as $7^{2\cdot 15}\cdot 7 > 8^{2\cdot 14}\cdot 8$. I also tried to write the right hand side as $\frac{8^{31}}{8^2}$.


Solution 1:

The following (not particularly elegant) proof uses reasonably basic multiplication and division.

We need to show that $7^{31} > 8^{29}$, i.e. that $\dfrac{7^{31}}{8^{29}}>1$.

We have: $\dfrac{7^{31}}{8^{29}}=\dfrac{7^{2}\cdot7^{29}}{8^{29}}=\dfrac{7^{3}}{8}\Big(\dfrac{7}{8}\Big)^{28}=\dfrac{7^{3}}{8}\Big(\dfrac{7^4}{8^4}\Big)^{7}=\dfrac{7^{3}}{8}\Big(\dfrac{2401}{4096}\Big)^{7} > \dfrac{7^{3}}{8}\Big(\dfrac{2400}{4100}\Big)^{7}=\dfrac{7^{3}}{8}\Big(\dfrac{24}{41}\Big)^{7}=\dfrac{7^{3}}{8}\dfrac{24}{41}\Big(\dfrac{24}{41}\Big)^{6}=\dfrac{3 \cdot 7^{3}}{41}\Big(\dfrac{24^2}{41^2}\Big)^{3}=\dfrac{3 \cdot 7^{3}}{41}\Big(\dfrac{576}{1681}\Big)^{3}>\dfrac{3 \cdot 7^{3}}{41}\Big(\dfrac{576}{1683}\Big)^{3}=\dfrac{3 \cdot 7^{3}}{41}\Big(\dfrac{9 \cdot 64} {9\cdot 187}\Big)^{3}=\dfrac{3 \cdot 7^{3}}{41}\Big(\dfrac{64} {187}\Big)^{3}=\dfrac{2^{18} \cdot 3 \cdot 7^3}{11^3 \cdot 17^3 \cdot 41}=\dfrac{2^{18} \cdot 3 \cdot 7^3}{1331 \cdot 17^3 \cdot 41}>\dfrac{2^{18} \cdot 3 \cdot 7^3}{1332 \cdot 17^3 \cdot 41}=\dfrac{2^{16} \cdot 7^3}{111 \cdot 17^3 \cdot 41}=\dfrac{2^{16} \cdot 7^3}{4551 \cdot 17^3}>\dfrac{2^{16} \cdot 7^3}{4557 \cdot 17^3}=\dfrac{2^{16} \cdot 7}{93 \cdot 17^3}=\dfrac{2^{16} \cdot 7}{1581 \cdot 17^2}>\dfrac{2^{16} \cdot 7}{1582 \cdot 17^2}=\dfrac{2^{15}}{113 \cdot 17^2}=\dfrac{2^{15}}{1921 \cdot 17}>\dfrac{2^{15}}{1924 \cdot 17}=\dfrac{2^{13}}{481 \cdot 17}=\dfrac{8192}{ 8177}>1. \quad\square$

Solution 2:

Others may bristle at this "proof," but:

$$7^{31} = 157,775,382,034,845,806,615,042,743 \\ 8^{29} = 154,742,504,910,672,534,362,390,528$$

If all else fails, just calculating the expressions and comparing them will work. This particular problem is only mildly tedious to attack this way if you have pen/paper.