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

Latest revision as of 10:37, 27 April 2008

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

$\mathrm{(A)}\ 7\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ 1\qquad\mathrm{(D)}\ 8\qquad\mathrm{(E)}\ 6$

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

@@ Line 2: / Line 2: @@
 If <math>2^x=15</math> and <math>15^y=256</math>, then the product <math>xy</math> equals
-A. <math>7</math>
+<math>\mathrm{(A)}\ 7\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ 1\qquad\mathrm{(D)}\ 8\qquad\mathrm{(E)}\ 6</math>
-B. <math>3</math>
-C. <math>1</math>
-D. <math>8</math>
-E. <math>6</math>
 ==Solution==