If the domain of functions f (x) = loga (x ^ 2-1) and GX = loga (x-1) + loga (x + 1) are f and g respectively, then their relationship is

If the domain of functions f (x) = loga (x ^ 2-1) and GX = loga (x-1) + loga (x + 1) are f and g respectively, then their relationship is

F：
x ²- 1 > 0 gives x > 1 or x < - 1
G：
X-1 ＞ 0 and X + 1 ＞ 0
Get x > 1
Relationship between F and G
G∈F

Given that a > 0 and a ≠ 1, the function y = loga (2x − 3)+ If P is on the image of power function f (x), then f (8) = ___

∵loga1=0，
When 2x-3 = 1, i.e. x = 2, y=
2，
The coordinate of point P is p (2,
2）．
Y = f (x) = Xa, because the image passes through the point (2,
2），
Get
2=2a，a=1
two
∴y=f（x）=x1
2，
f（8）=81
2＝2
two
So the answer is: 2
2．

It is known that the image of a > 0 and a ≠ 0 function y = ㏒ a (2x-3) + root 2 passes through the fixed point P. if P is on the power function image, then f (8)=

Because the logarithm of 1 is 0
The image of function y = ㏒ a (2x-3) + root 2 passes through the fixed point (2, √ 2)
Let the power function be y = x ^ a
∴ √2=2^a
∴ a=1/2
∴ f(x)=x^(1/2)=√x
∴ f(8)=√8=2√2

If the function f (x) = loga (x + radical x ^ 2 + 2A ^ 2) is an odd function, then a =? process

f(-x)=loga[-x+√(x^2+2a^2)]
=-f(x)=-loga[x+√(x^2+2a^2)]
=loga{1/[x+√(x^2+2a^2)]}
So - x + √ (x ^ 2 + 2A ^ 2) = 1 / [x + √ (x ^ 2 + 2A ^ 2)]
So [x + √ (x ^ 2 + 2A ^ 2)] [- x + √ (x ^ 2 + 2A ^ 2)] = 1
So x ^ 2 + 2A ^ 2-x ^ 2 = 1
a^2=1/2
A is the base greater than 0
a=√2/2

If the image of the function y = loga (2x-3) + 4 passes through the fixed point m and the point m is on the image of the power function f (x), then f (3) = __

∵loga1=0，
When 2x-3 = 1, i.e. x = 2, y = 4,
The coordinates of point m are p (2, 4)
Power function f (x) = x α The image passing point m (2, 4),
So 4 = 2 α， Solution α= 2；
So the power function is f (x) = x2
Then f (3) = 9
So the answer is: 9

The image of function f (x) = AX2 + BX + A-3 is symmetrical about the Y axis, and its definition domain is [A-4, a] (a, B belong to R) to find the value domain of F (x)

The image of function f (x) = AX2 + BX + A-3 is symmetrical about the Y axis, that is, the even function f (- x) = ax2-bx + A-3 = f (x), so B = 0f (x) = AX2 + A-3. Its definition field is [A-4, a] (a, B ∈ R). Because the definition field should be symmetrical about the origin, A-4 = - AA = 2F (x) = 2x ^ 2-1, f (x) = - 1, Max x = 2 or F (x)