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> | + | 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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<math>\text{(A) } 12\sqrt{2}\quad | <math>\text{(A) } 12\sqrt{2}\quad | ||
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== Solution == | == Solution == | ||
− | <math>\fbox{B}</math> | + | 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
Solution
Rewrite the first equation as and this is of the form ; by inspection it is easy to see that or . Therefore (or vice versa—it doesn't matter here) so . Substituting this into , this means and , so which is
See also
1990 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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All AHSME Problems and Solutions |
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