f ( x y ) = f ( x ) + f ( y ) f(xy)=f(x)+f(y)

Algebra Level 1

For positive real numbers x x and y y , function f f satisfies f ( x y ) = f ( x ) + f ( y ) , f ( 2 ) = 16. f(xy)=f(x)+f(y), f(2)=16. What is the value of f ( 8 ) + f ( 1 2 ) ? f(8)+f\left(\frac{1}{2}\right)?

32 24 40 48

Ajay Dutta by Ajay Dutta , 24, India

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2 solutions

Note that f ( x y z ) = f ( x y ) + f ( z ) = f ( x ) + f ( y ) + f ( z ) f(xyz) = f(xy) + f(z) = f(x) + f(y) + f(z)

Therefore, f ( x 3 ) = 3 f ( x ) f(x^3) = 3f(x)

Also, f ( 1 ) = f ( 1 ) + f ( 1 ) → f ( 1 ) = 0 f(1) = f(1) + f(1) \rightarrow f(1) = 0

And, f ( 1 ) = f ( x ) + f ( 1 x ) → f ( x ) = − f ( 1 x ) f(1) = f(x) + f(\frac{1}{x}) \rightarrow f(x) = -f(\frac{1}{x})

Now, f ( 2 3 ) + f ( 1 2 ) = 3 f ( 2 ) − f ( 2 ) = 2 f ( 2 ) = 32 f(2^3) + f(\frac{1}{2}) =3f(2) - f(2) = 2f(2) = \boxed{32}

3 Helpful 1 Interesting 1 Brilliant 0 Confused
Shashi Yadav
Feb 28, 2014

since f(xy)=f(x)+f(y), & f(2)=16, therefore, f(8)+f(1/2)=f(8.1/2) =f(2.2) =f(2)+f(2) =16+16 =32 #

2 Helpful 0 Interesting 0 Brilliant 0 Confused

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