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 $\fbox{B}$

1990 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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1990 AHSME Problems/Problem 23

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