Math Archive

Algebra Level 3

Consider a real valued function f ( x ) f(x) satisfying , 2 f ( x y ) = ( f ( x ) ) y + ( f ( y ) ) x 2f(xy)=(f(x))^{y}+(f(y))^{x} for all x , y x,y belonging to the set of real numbers . And f ( 1 ) = 2 f(1)=2 , then find ( ∑ r = 1 15 f ( r ) ) + 1000 (\sum_{r=1}^{15} f (r))+1000


The answer is 66534.

Shivam Jadhav by Shivam Jadhav , 22, India

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2 solutions

From 2 f ( x y ) = ( ( f ( x ) ) y + ( f ( y ) ) x \quad 2f(xy) = ((f(x))^y + (f(y))^x

⇒ 2 f ( 1 ˙ y ) = ( f ( 1 ) ) y + ( f ( y ) ) 1 2 f ( y ) = 2 y + f ( y ) ⇒ f ( y ) = 2 y ∑ r = 1 15 f ( r ) + 1000 = ∑ r = 1 15 2 r + 1000 = 2 16 − 1 2 − 1 − 1 + 1000 = 66534 \begin{aligned} \Rightarrow 2f(1\dot{}y) & = (f(1))^y + (f(y))^1 \\ 2f(y) & = 2^y + f(y) \\ \Rightarrow f(y) & = 2^y \\ \sum_{r=1}^{15} {f(r)} + 1000 & = \sum_{r=1}^{15} {2^r} + 1000 \\ & = \frac{2^{16}-1}{2-1} -1 + 1000 \\ & = \boxed{66534} \end{aligned}

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Rohan Shah
May 6, 2015

Just take f(2)= 2X1; f(3) = 3X1...... so by substitute in 2f(xy) = f(x) ^y + f(y) ^x so, y = 1 and x = 2,3,4.....15 hence you will get; f(1) + f(2)+....+f(15)= 2 + 2^2 + 2^3 +....+ 2^15. As it forms a G.P.; summation equals to 65534 and add 1000.... so; 66534

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