$2^{50} < 3^{32}$ using elementary number theory

Solution 1:

Compare: $$3^{32}=(3^{2})^{16}\quad\text{vs.}\quad2^{50}=4(2^{3})^{16}$$ So that using the binomial theorem to second order: $$\frac{3^{32}}{2^{50}}=\frac{(9/8)^{16}}{4}=\frac{(1+1/8)^{16}}{4} >\frac{1+16/8+120/64}{4}>1$$

Solution 2:

$2^{11} = 2048 < 2187 = 3^7$ and so $\log_3 2 < 7/11$.

Thus, $50 \log_3 2 < 50\cdot 7/11 < 32$, which implies the result.

Proving that $x=50$ is the largest integer solution for $2^x < 3^{32}$ is harder. You need to know that $2^{27} > 3^{17}$ to get $x \le 50$. It all boils down to approximating $\log_3 2 = 0.630929\cdots$ by the ratio of small integers numbers.

Solution 3:

My approach was similar to that of nbubis, namely, after brutal root extraction, proving

that $\sqrt[\Large8]2<\dfrac98~.~$ Now, $\sqrt2=\sqrt{\dfrac{100}{50}}<\sqrt{\dfrac{100}{49}}=\dfrac{10}7~.~$ Then we have $\sqrt[\Large4]2=\sqrt{\sqrt2}<\sqrt{\dfrac{10}7}$

$=\sqrt{\dfrac{100}{70}}<\sqrt{\dfrac{100}{64}}=\dfrac{10}8=\dfrac54~.~$ Lastly, $\sqrt[\Large8]2=\sqrt{\sqrt[\Large4]2}<\sqrt{\dfrac54}=\sqrt{\dfrac{80}{64}}<\sqrt{\dfrac{81}{64}}=\dfrac98~.$

Solution 4:

Note first that this is equivalent to $2^{25}\lt 3^{16}$ or $2\cdot 2^{24}\lt 3\cdot 3^{15}$ or $2\cdot 4^3\cdot 2^{24}\lt 3\cdot 4^3 \cdot 3^{15}$

Now $4\cdot 2^8=1024=10^3+24$ and $4\cdot 3^5=972=10^3-28$

So we can rewrite the last form of the inequality as equivalent to $$2\cdot(10^3+24)^3\lt 3\cdot (10^3-28)^3$$

Expanding the factors gives $$2\cdot 10^9+6\cdot24\cdot 10^6+6\cdot24^2\cdot10^3+24^3\lt 3\cdot 10^9-9\cdot28\cdot10^6+9\cdot 28^2\cdot10^3-28^3$$

Rearranging, this gives:

$$10^9+(9\cdot 28^2-6\cdot24^2)\cdot10^3\gt(9\cdot28+6\cdot24)\cdot10^6+28^3+24^3$$

Now it is obvious that this is true, because $9\cdot 28+6\cdot 24\lt 300+150\lt 500$