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

(Created page with "== Problem == If <math>x,y>0, log_y(x)+log_x(y)=\frac{10}{3} \text{ and } xy=144,\text{ then }\frac{x+y}{2}=</math> <math>\text{(A) } 12\sqrt{2}\quad \text{(B) } 13\sqrt{3}\qu...")
 
 
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== Problem ==
 
== Problem ==
  
If <math>x,y>0, log_y(x)+log_x(y)=\frac{10}{3} \text{ and } xy=144,\text{ then }\frac{x+y}{2}=</math>
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If <math>x,y>0, \log_y(x)+\log_x(y)=\frac{10}{3} \text{ and } xy=144,\text{ then }\frac{x+y}{2}=</math>
 
 
  
 
<math>\text{(A) } 12\sqrt{2}\quad
 
<math>\text{(A) } 12\sqrt{2}\quad
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== Solution ==
 
== Solution ==
<math>\fbox{B}</math>
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Rewrite the first equation as <cmath>\frac{\log y}{\log x}+\frac{\log x}{\log y}=3\tfrac13</cmath> and this is of the form <math>u+\tfrac1u =3\tfrac13</math>; by inspection it is easy to see that <math>u=3</math> or <math>\tfrac13</math>. Therefore <math>\log y=3\log x</math> (or vice versa—it doesn't matter here) so <math>y=x^3</math>. Substituting this into <math>xy=144</math>, this means <math>x=2\sqrt{3}</math> and <math>y=24\sqrt{3}</math>, so <math>\tfrac12(x+y)=13\sqrt{3}</math> which is <math>\fbox{B}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 12:41, 4 February 2016

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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