# Difference between revisions of "1990 AHSME Problems/Problem 23"

## Problem

If $x,y>0, \log_y(x)+\log_x(y)=\frac{10}{3} \text{ and } xy=144,\text{ then }\frac{x+y}{2}=$

$\text{(A) } 12\sqrt{2}\quad \text{(B) } 13\sqrt{3}\quad \text{(C) } 24\quad \text{(D) } 30\quad \text{(E) } 36$

## Solution

Rewrite the first equation as $$\frac{\log y}{\log x}+\frac{\log x}{\log y}=3\tfrac13$$ and this is of the form $u+\tfrac1u =3\tfrac13$; by inspection it is easy to see that $u=3$ or $\tfrac13$. Therefore $\log y=3\log x$ (or vice versa—it doesn't matter here) so $y=x^3$. Substituting this into $xy=144$, this means $x=2\sqrt{3}$ and $y=24\sqrt{3}$, so $\tfrac12(x+y)=13\sqrt{3}$ which is $\fbox{B}$