Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 10"

Revision as of 19:17, 17 October 2007

If $2^x=15$ and $15^y=256$ , then the product $xy$ equals

A. $7$

B. $3$

C. $1$

D. $8$

E. $6$

$(2^x)^y = 2^{xy} = 256$ , so $xy = 8\ \mathrm{(D)}$ .

@@ Line 1: / Line 1: @@
 ==Problem==
-{{empty}}
+If <math>2^x=15</math> and <math>15^y=256</math>, then the product <math>xy</math> equals
+A. <math>7</math>
+B. <math>3</math>
+C. <math>1</math>
+D. <math>8</math>
+E. <math>6</math>
 ==Solution==
-{{solution}}
+<math>(2^x)^y = 2^{xy} = 256</math>, so <math>xy = 8\ \mathrm{(D)}</math>.
 ==See also==
 {{CYMO box|year=2006|l=Lyceum|num-b=9|num-a=11}}
+[[Category:Introductory Algebra Problems]]