1990 AHSME Problems/Problem 23

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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 $\fbox{B}$

See also

1990 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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