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2006 Cyprus MO/Lyceum/Problem 10

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Problem

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

Solution

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

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 9
Followed by
Problem 11
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