The domain of ∵ function is (0, +∞).

∴a^x-kb^x>； The solution set of 0 is exactly (0, +∞), (note the key word: exactly)

When x=0, an x-kb x = 0-kb 0 = 0,

That is, 1-k=0, ∴k= 1

Therefore, f (x) = ln (a x-b x), (a>1> b & gt0);

The function f(x) takes a positive value in (1, +∞).

∴f( 1)=0，

That is ln(a-b)=0,

∴a-b= 1,a=b+ 1，

f(x)=ln[(b+ 1)^x-b^x],(0<； b & lt 1);

∫f(3)= ln4，

∴ln[(b+ 1)^3-b^3]=ln4

(b+ 1)^3-b^3=4

(b^3+3b^2+3b+ 1)-b^3=4

3b^2+3b-3=0

b^2+b- 1=0

To solve a quadratic equation with one variable,

b=(- 1 √5)/2，

∫0 & lt； b & lt 1,

∴b=(√5- 1)/2，

At this time a=b+ 1=(√5+ 1)/2,

So there are real numbers a=(√5+ 1)/2 and b=(√5- 1)/2.