Answer:
False
Step-by-step explanation:
We can disprove this by plugging in u and v for f(x) = lnx
f(u) = 2f(v)
ln(u) = 2ln(v)
[tex]ln(u) = ln(v^2)[/tex]
[tex]u = v^2[/tex]
which is only true if u = v = 1 or 0
If u = e and v = [tex]e^2[/tex]
f(u) = f(e) = lne = 1
[tex]f(v) = f(e^2) = lne^2 = 2
So f(v) = 2f(u). But not the other way around