Given the functions f (x) = loga (x) and G (x) = 2loga (2x + T-2), (a > 0, a ≠ 1, t ∈ R), the tangents of the image at x = 2 are parallel to each other (1) Finding the value of T (2) Let f (x) = g (x) - f (x), when x ∈ [1,4], f (x) ≥ 2 is constant, and the value range of a is obtained Note: A is the log base

Given the functions f (x) = loga (x) and G (x) = 2loga (2x + T-2), (a > 0, a ≠ 1, t ∈ R), the tangents of the image at x = 2 are parallel to each other (1) Finding the value of T (2) Let f (x) = g (x) - f (x), when x ∈ [1,4], f (x) ≥ 2 is constant, and the value range of a is obtained Note: A is the log base

f(x)'=1/(x*lna)
g(x)'=(2*2)/[(2x+t-2)*lna]
When x = 2, f (x) '= 1 / 2lna
g(x)'=4/(2+t)lna
So 1 / 2 = 4 / (2 + T), t = 6
g(x)=2loga(2x+4)
F(x)=2loga(2x+4)-loga(x)=2loga[(2x+4)/x]=2loga[2+(4/x)]
Because in X ∈ [1,4], 3 ≤ 2 + (4 / x) ≤ 6
It is also established that f (x)) ≥ 2, a > 1
So loga (T) is an increasing function
Only the minimum value of loga (T) is greater than or equal to 2, and the minimum value is obtained at 2 + (4 / x) = 3
So 2loga (3) ≥ 2, loga (3) ≥ loga (a)
So a ≤ 3
To sum up, 1