Competition problem — Algebra — Maths Olympiad Prep

Library / /1 of 34

Algebra Difficulty 4.6 AIME Prove it Romania

a) Compare the numbers $2^{53}$ and $3^{35}$ .

b) Show that, if $5b \ge 3a > 0$ , then $2^{ab} < 3^{ba}$ .

Solution

a) Since $3^7 = 2187 > 2048 = 2^{11}$ , we find $3^{35} = (3^7)^5 > (2^{11})^5 > 2^{53}$ .

b) The inequality $2^{53} < 3^{35}$ can be written $\left(\frac{2^{10}}{3}\right)^5 < \left(\frac{3^{10}}{2}\right)^3$ . It follows that $\left(\frac{2^{10}}{3}\right)^{5a} < \left(\frac{3^{10}}{2}\right)^{3a}$ . The last inequality implies $\left(\frac{2^{10}}{3}\right)^{5a} < \left(\frac{3^{10}}{2}\right)^{5b}$ or $\left(\frac{2^{10}}{3}\right)^a < \left(\frac{3^{10}}{2}\right)^b$ , hence $2^{10a+b} < 3^{10b+2}$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.