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If the function f (x) = log defined in (- 1,0) takes 2A as the base and X + 1 as the true number If f (x) is greater than 0, the value range of a is
f(x)=log2a(x+1)
Because the definition field is (- 1,0)
So (x + 1) ∈ (0,1)
So 0
If the logarithm of the function f (x) = log defined in the interval (- 1,0) with 2A as the base (x + 1) satisfies f (x) > 0, find the value range of a.tle The process should be complete
Because x ∈ (- 1,0), x + 1 ∈ (0,1), i.e. 0 < x + 1 < 1, so f (x) = log with 2A as the base (x + 1) logarithm > 0 = log with 2A as the base 1 logarithm. It can be seen that f (x) is a subtractive function in the definition field, so the base 2A ∈ (0,1), i.e. 0 < 2A < 1, so 0 < a < 1 / 2, i.e. a ∈ (0,1 / 2)
The monotone function f (x) defined on R satisfies that f (3) = log base is 2, logarithm is 3, and f (x + y) = f (x) + F (y) for any X and Y belonging to R, Prove that f (x) is an odd function; If f (k * 3 ^ x) + F (3 ^ X-9 ^ X-2) < 0 is constant for any real number, find the value range of K?
If y = - x is taken to get f (x) + F (- x) = 0, it is an odd function. Because f (0) = 0 < f (3) = log3 (2) f (x) is monotonous, f (x + y) = f (x) + f (y) = = = > > F (k * 3 ^ x) + F (3 ^ X-9 ^ X-2) = f (k * 3 ^ x + 3 ^ X-9 ^ X-2) 0. If z = 3 ^ x, it only needs Z ²- (K + 1) Z + 2 = 0 has two negative
If f (x) = the logarithm of (x + 2) with (2a-3) as the base, the logarithm of {log (2a-3) is (x + 2)} satisfies f (x)
2a-3 this is necessary
Because this logarithm satisfies f (x) < 0
So 2a-3 > 1
So a > 2
Given that the domain of the function y = log2 (x ^ 2 + KX + 3) is a real number set R, the value range of Changshu K is A、R B、（-2√3,2√3） C（-∞,-2√3]∪[2√3,+∞) D-empty set
So whatever value x takes
Both x ^ 2 + KX + 3 are greater than 0
So discriminant
Given that the domain of the function y = log2 (x ^ 2 + KX + 3) is a real number set R, the value range of the constant k?
The domain of function y = log2 (x ^ 2 + KX + 3) is the set of real numbers R, that is, x ^ 2 + KX + 3 > 0 is constant for all R
So there is a discriminant = k ^ 2-12
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