That's cheating. You're not solving, you're guessing that f(x) = a*ln(x) because you know the answer. Without guessing anything and supposing that I'd never heard about logarithms because that day it was raining and I missed that class, we can solve this functional equation by doing this: f(x^y) = y*f(x) if x = 1, f(1) = y*f(1) then f(1) = 0 differentiating with respect of x, f'(x^y)*y*x^(y-1) = y*f'(x) if y = 0, f'(1)/x = f'(x) finally f(x) = integralbetween0andxof[(f'(1)/t*dt]
f(x^y) = y f(x) Let x = e. Then f(e^y) = y f(e). Now, let t = e^y. Then y = ln(t). So, f(t) = ln(t) f(e). Note that f(e) is a constant, so we'll call it c. So f(t) = c ln(t). Or f(x) = c ln(x).
That's cheating. You're not solving, you're guessing that f(x) = a*ln(x) because you know the answer.
Without guessing anything and supposing that I'd never heard about logarithms because that day it was raining and I missed that class, we can solve this functional equation by doing this:
f(x^y) = y*f(x)
if x = 1, f(1) = y*f(1) then f(1) = 0
differentiating with respect of x,
f'(x^y)*y*x^(y-1) = y*f'(x)
if y = 0,
f'(1)/x = f'(x)
finally f(x) = integralbetween0andxof[(f'(1)/t*dt]
Let z be > 0.
f(z) = f(e^log(z))
Apply the functional equation with x = e and y = log(z) which gives:
f(z) = log(z) f(e).
separation: f(z)/ln(z) = f(x)/ln(x) = konst.
f(x^y) = y f(x)
Let x = e. Then f(e^y) = y f(e). Now, let t = e^y. Then y = ln(t). So, f(t) = ln(t) f(e). Note that f(e) is a constant, so we'll call it c. So f(t) = c ln(t). Or f(x) = c ln(x).
how can you just put a value of x which is not a variable
@@abhirupkundu2778 Because it's true for any positive value, so that means it's true for x = e.