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) and the value of T is obtained

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) and the value of T is obtained

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 also holds that f (x)) ≥ 2 is constant
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 & lt; a ≤ 3

Given the functions f (x) = 2loga (x) and G (x) = loga (2x + T-2), (a > 0, a ≠ 1, t ∈ R), the tangents of the image at x = 2 are parallel to each other (1) and the value of T is obtained

f'(x)=2/xlna g'(x)=2/(2x+t-2)lna f'(2)=g'(2)
The solution is t = 0

Given the function f (x) loga (x + 1), G (x) = 2loga (2x + T) (t ∈ R), where x ∈ [0,15]. A > 0, a ≠ 1
(1) If 1 is a solution of the equation f (x) = g (x) about X, find the value of T
(2) When 0 ＜ a ＜ 1, the inequality f (x) ≥ g (x) holds and the value range of T is obtained
I am a senior one. Please give me more advice

Given the function f (x) = Log &; a &; (x + 1), G (x) = 2log &; a &; (2x + T) (t ∈ R), where x ∈ [0,15]. A ＞ 0, a ≠ 1. (1) if 1 is a solution of the equation f (x) = g (x) about X, find the value of T. (2) when 0 ＜ a ＜ 1, the inequality f (x) ≥ g (x) holds, find the value of T