The function \( f \) satisfies the functional equation \( f(xy) = \frac{f(x)}{y} \) for all \( x, y > 0 \). Given that \( f(500) = 3 \), the value of \( f(600) \) can be calculated by substituting \( x = 500 \) and \( y = \frac{600}{500} = \frac{6}{5} \). This leads to \( f(600) = \frac{f(500)}{\frac{6}{5}} = \frac{3}{\frac{6}{5}} = \frac{5}{2} \). Therefore, \( f(600) = 2.5 \).
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Is this a trick question or did I just stumble upon a quick way to do this? I usually go insane when you post one of these.Albert said:given a function $f$, satisfying :
(1) $f(xy)=\dfrac {f(x)}{y}$ for all $x,y>0$
(2) $f(500)=3$
what is the value of $f(600)$
please don't go insane,your answer is correct,thanks for participating.topsquark said:Is this a trick question or did I just stumble upon a quick way to do this? I usually go insane when you post one of these.
[sp]
[math]f(xy) = \frac{f(x)}{y}[/math]
Let x = 500 and xy = 600. Then y = 600/500 = 6/5.
Thus
[math]f(600) = \frac{f(500)}{\frac{6}{5}} = \frac{3}{\frac{6}{5}} = \frac{5}{2}[/math]
[/sp]
-Dan
I don't usually lack self-confidence but whenever I can quickly answer a problem from you or the POTWs I get the feeling I've overlooked a critical point! Like this one...I wasn't expecting a one line solution so I was unsure about it. Personally I prefer the challenge, though I usually don't post my attempts. I like these. (Yes)Albert said:please don't go insane,your answer is correct,thanks for participating.
By the way ,do you prefer more challenging problems (need a lot of tricks)?