Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 10"
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==Problem== | ==Problem== | ||
| + | If <math>2^x=15</math> and <math>15^y=256</math>, then the product <math>xy</math> equals | ||
| + | |||
| + | <math>\mathrm{(A)}\ 7\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ 1\qquad\mathrm{(D)}\ 8\qquad\mathrm{(E)}\ 6</math> | ||
==Solution== | ==Solution== | ||
| − | {{ | + | <math>(2^x)^y = 2^{xy} = 256</math>, so <math>xy = 8\ \mathrm{(D)}</math>. |
==See also== | ==See also== | ||
{{CYMO box|year=2006|l=Lyceum|num-b=9|num-a=11}} | {{CYMO box|year=2006|l=Lyceum|num-b=9|num-a=11}} | ||
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| + | [[Category:Introductory Algebra Problems]] | ||
and 
, then the product 
equals
, so 
.
