(1) For any real number a of inequality 1, the fixed point of the image constant of the function f (x) = loga (x + 3) - 2x is (2) Let the image of inverse function of function f (x) = loga (x + 2) + 1 (a > 0, a ≠ 1) pass through points (3,4), and find the value of A

(1) For any real number a of inequality 1, the fixed point of the image constant of the function f (x) = loga (x + 3) - 2x is (2) Let the image of inverse function of function f (x) = loga (x + 2) + 1 (a > 0, a ≠ 1) pass through points (3,4), and find the value of A

1. By a ^ (y + 2x) = (x + 3), and a ^ 0 = 1
By solving the equations y + 2x = 0, x + 3 = 1, x = - 2, y = 4
Therefore, the image is always over the fixed point (- 2,4)
2. If the image of the inverse function passes through points (3,4), then the original function passes through points (4,3)
So there is: 3 = loga (6) + 1
loga (6)=2
a=√6