If the definition field of function f (x) is {XLX ＞ 0}, and f (x) is an increasing function, f (x · y) = f (x) + F (y). F (3) = 1 And f (a) > F (A-1) + 2

If the definition field of function f (x) is {XLX ＞ 0}, and f (x) is an increasing function, f (x · y) = f (x) + F (y). F (3) = 1 And f (a) > F (A-1) + 2

f（a-1）+2=f（a-1）+1+1=f（a-1）+f(3)+f(3)=f(3*3*(a-1))=f(9a-9)
Increasing function and domain x > 0
therefore
a>0
a-1>0
a>9a-9
one

If f (x) is an increasing function in the domain (0. + ∞), tangent to all x, Y > 0, f (x / y) = f (x) - f (y),
If f (6) = 1, f (x + 3) - f (1 / 3)

Because f (Y / x) = f (x) - f (y)
So f (x + 3) - f (1 / 3) = f (3 (x + 3)) 0 and 3 (x + 3)

For any value of X in the domain of function y = f (x), f (x + a) = f (A-X). This paper proves that the image of y = f (x) is symmetric with respect to the line x = a

∵f(x+a)=f(a-x)
Replace X in the above formula with x-a to obtain
f(x)=f(2a-x)
∵ x + (2a-x) = 2 * a, f (x) = f (2a-x), that is, the ordinates are the same
The point with X and 2a-x as abscissa is symmetric with respect to the line x = a
These two points are arbitrary and always on the image of y = f (x)
The image of y = f (x) is symmetric with respect to the line x = a