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

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==Problem==
 
==Problem==
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If <math>2^x=15</math> and <math>15^y=256</math>, then the product <math>xy</math> equals
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A. <math>7</math>
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B. <math>3</math>
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C. <math>1</math>
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D. <math>8</math>
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E. <math>6</math>
  
 
==Solution==
 
==Solution==
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<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]]

Revision as of 20:17, 17 October 2007

Problem

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

A. $7$

B. $3$

C. $1$

D. $8$

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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30