Difference between revisions of "2011 AIME I Problems/Problem 9"

Revision as of 11:50, 19 March 2011

Problem

Suppose $x$ is in the interval $[0, \pi/2]$ and $\log_{24\sin x} (24\cos x)=\frac{3}{2}$ . Find $24\cot^2 x$ .

Solution

We can rewrite the given expression as $\sqrt{24^3\sin^3 x}=24\cos x$ . Square both sides and divide by $24^2$ to get $24\sin ^3 x=\cos ^2 x$ Rewrite $\cos ^2 x$ as $1-\sin ^2 x$ $24\sin ^3 x=1-\sin ^2 x$ $24\sin ^3 x+\sin ^2 x - 1=0$ Testing values using the rational root theorem gives $\sin x=\frac{1}{3}$ as a root. $\Arcsin \frac{1}{3}$ (Error compiling LaTeX. Unknown error_msg) does fall in the first quadrant so it satisfies the interval. Thus $\sin ^2 x=\frac{1}{9}$ . Using the Pythagorean Identity gives us $\cos ^2 x=\frac{8}{9}$ . Then we use the definition of $\cot ^2 x$ to compute our final answer. $24\cot ^2 x=24\frac{\cos ^2 x}{\sin ^2 x}=24\left(\frac{\frac{8}{9}}{\frac{1}{9}}\right)=24(8)=\boxed{192}$ .

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 == Problem ==
-Suppose <math>x</math> is in the interval <math>[0, \pi/2]</math> and <math>\log_(24\sin x) (24\cos x)=\frac{3}{2}</math>. Find <math>24\cot^2 x</math>.
+Suppose <math>x</math> is in the interval <math>[0, \pi/2]</math> and <math>\log_{24\sin x} (24\cos x)=\frac{3}{2}</math>. Find <math>24\cot^2 x</math>.
 == Solution ==

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Difference between revisions of "2011 AIME I Problems/Problem 9"

Revision as of 11:50, 19 March 2011

Problem

Solution