THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

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 $2^{10a+b}<3^{10b+2}$.