Let the domain of the function y = f (x) be (0, + infinity), and f (XY) = f (x) + F (y) be constant for any positive real number x, y, and f (2) = 1, and 1. Find f (8) 2. Solve inequality f (x) + F (X-2) < = 3

Let the domain of the function y = f (x) be (0, + infinity), and f (XY) = f (x) + F (y) be constant for any positive real number x, y, and f (2) = 1, and 1. Find f (8) 2. Solve inequality f (x) + F (X-2) < = 3

The function f (x) is an increasing function defined on (0, ∞), and f (XY) = f (x) + F (y) is constant for any positive real number x, y, and f (2) = 1,
1. Find f (8)
2. Solve inequality f (x) + F (X-2)

If f (XY) = f (x) × F (y) holds for all real numbers x and y, and f (0) is not equal to 0, then f (2005)=______

Let x = y = 0, from F (XY) = f (x) × F (0) = f (0) from F (y) × 0)=f(0)*f(0)=[f(0)]^2f(0) × [1-f (0)] = 0 because f (0) ≠ 0, f (0) = 1. Let x = 2005, y = 0, get f (2005) × 0)=f(0)=f(2005) × F (0), i.e. 1 = f (2005) × 1 so f (2005) = 1. In fact, in this example, let y = 0 to get f (x) × 0)=f(0...

If f (XY) = f (x) * f (y) holds for all real numbers and f (0) is not equal to 0, then the value of F (2005) is

F (2005) f (0) = f (2005 * 0) = f (0). Because f (0) is not equal to 0, both sides are eliminated to get f (2005) = 1

If f (XY) = f (x) * f (y) holds for all real numbers x and y, and f (0) is not equal to 0, what is f (2009) equal to

Let y = 0 and get f (x * 0) = f (0) = f (x) * f (0)
Because f (0) ≠ 0, then f (x) = 1
So f (2009) = 1

If f (XY) = f (x) * f (y) holds for all real numbers x and y, and f (0) is not equal to 0, what is f (2009) equal to? If you know, please tell me, thanks

Let x = 0 and y = 2009, then f (XY) = f (0) = f (0) x f (2009)
Because f (0) is not equal to 0, both sides divide by F (0)
F (2009) = 1

If f (XY) = f (x) * f (y) holds for all real numbers x and y, and f (0) 0, then f (2005)=

Are you missing f (0) ≠ 0, or you won't do it
If omitted, then
Let y = 0 to obtain
f(x × 0)=f(0)=f(x) × f(0)
That is, 1 = f (x) × one
F (x) = 1, that is, f (x) is a constant function 1. Therefore, f (2005) = 1