2007 Cyprus MO/Lyceum/Problem 2: Difference between revisions

Latest revision as of 19:30, 6 May 2007

Given the formula $f(x) = 4^x$ , then $f(x+1)-f(x)$ equals to

$\mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 4^x\qquad \mathrm{(C) \ } 2\cdot4^x\qquad \mathrm{(D) \ } 4^{x+1}\qquad \mathrm{(E) \ } 3\cdot4^x$

$f(x+1)-f(x)=4^{x+1}-4^x=4\cdot4^x-4^x=(4-1)\cdot4^x=3\cdot4^x\Longrightarrow\mathrm{ E}$

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 ==Solution==
-<math>f(x+1)-f(x)=4^{x+1}-4^x=4\cdot4^x-4^x=4^x\cdot(4-1)=3\cdot4^x\Rightarrow\mathrm{ E}</math>
+<math>f(x+1)-f(x)=4^{x+1}-4^x=4\cdot4^x-4^x=(4-1)\cdot4^x=3\cdot4^x\Longrightarrow\mathrm{ E}</math>
 ==See also==
-*[[2007 Cyprus MO/Lyceum/Problems]]
+{{CYMO box|year=2007|l=Lyceum|num-b=1|num-a=3}}
-*[[2007 Cyprus MO/Lyceum/Problem 1|Previous Problem]]
+[[Category:Introductory Algebra Problems]]
-*[[2007 Cyprus MO/Lyceum/Problem 3|Next Problem]]