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

Revision as of 11:45, 19 March 2011

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

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 $24\sin ^3 x=\cos ^2 x$

@@ Line 1: / Line 1: @@
 == 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 ==
+We can rewrite the given expression as
+<math>\sqrt{24^3\sin^3 x}=24\cos x</math>.
+Square both sides and divide by <math>24^2</math> to get
+<math>24\sin ^3 x=\cos ^2 x</math>