It is known that the function f (x) = log a 1-mx / X-1 is odd (a is greater than 0, and a ≠ 1) (1) Find the value of M; (2) The monotonicity of F (x) on interval (1, + ∞) is judged and proved; (3) When a ＞ 1, X ∈ (1, √ 3), the range of F (x) is (1, + ∞);

It is known that the function f (x) = log a 1-mx / X-1 is odd (a is greater than 0, and a ≠ 1) (1) Find the value of M; (2) The monotonicity of F (x) on interval (1, + ∞) is judged and proved; (3) When a ＞ 1, X ∈ (1, √ 3), the range of F (x) is (1, + ∞);

(1) F (- x) = - f (x) so log a (1 + MX) / (- 1-x) = log a (x-1) / (1-mx) so (1 + MX) / (- 1-x) = (x-1) / (1-mx) the solution is m = - 1 (2) x > 1F (x) = log a (1 + x) - log a (x-1) f '(x) = ((x + 1) LNA) ^ - 1) - ((x-1) LNA) ^ - 1) f' (x) = - 2 / (LNA (x ^ 2-1)) if 0

Let f (x) be an odd function, and when x > 0, f (x) = log12x (I) find the analytic expression of F (x) when x < 0; (II) solve the inequality f (x) ≤ 2

(I) let x ＜ 0, then − x ＞ 0 {f (− x) = log12 (− x)} f (x) = − f (− x) = log12 (− x). So: when x ＜ 0, f (x) = - Log & nbsp; 12 (- x). (II) from the meaning of the title, we get x ＞ 0 log12x ≤ 2 or X ＜ 0 − log12 (− x) ≤ 2 {x ≥ 14 or − 4 ≤ x ＜ 0

Given that the function f (x) = log | x 1 | with base a has f (x) greater than 0 in the interval (- 1,0), then the following conclusion is correct:
1. F (x) is an increasing function on (negative infinity, 0)
2. F (x) is a decreasing function on (negative infinity, 0)
On the increasing function, f is infinite
4. F (x) is a decreasing function on (negative infinity, - 1)
A plus sign is missing:
If we know that the function f (x) = log | x + 1 | with the base of a has f (x) greater than 0 on the interval (- 1,0),

When x ∈ (- 1,0), | x + 1 | ∈ (0,1), that is, the true number is a positive pure decimal, and the logarithm is greater than 0, which indicates that the image of base a ∈ (0,1) and function y = f (x) is an axisymmetric figure with the axis of line x = - 1. The figure on the right side of the line has the same shape as the logarithm function image of base a ∈ (0,1), which is a decreasing function, and the left side is an increasing function

Let 3A + 4B = AB be the minimum of a + B

Is log2a + log2b = log2 (a + b)?

It's not equal
log(a)(MN)=log(a)(M)+log(a)(N);
log2A+log2B=log2(AB)

Given that 0 < a < 10 < B < 1 and log2a × log2b, the maximum value of log2 (AB) is 0

Because 0 < a < 1, 0 < B < 1,
So log2a < 0, log2b < 0
（-log2a）＞0,（-log2b）＞0
So (- log2a) + (- log2b) ≥ 2 √ [(- log2a) (- log2b)] = 2 √ [log2a * log2b] = 2 √ 16 = 8
So - [(- log2a) + (- log2b)] ≤ - 8
Log2 (AB)
=log2a+log2b
=-[（-log2a）+（-log2b）]≤-8,
Then the maximum value of log2 (AB) is - 8
If satisfied, please accept! Thank you

If log2a + log2b = 6, then the minimum value of a + B is ()
A. 26B. 6C. 82D. 16

∵ log2a + log2b = log2ab = 6, ∵ 26 = AB, ∵ a + B ≥ 2Ab = 16, if and only if a = B, the equal sign is taken, that is, the minimum value of a + B is 16, so D is selected

That is: a > 1, b > 1, log2a x log2b = 4, what is the minimum value of AB?
We need it now

4=log2 a×log2 b≤[(log2 a+log2 b)/2]^2=(log2 √ab)^2
And a > 1, b > 1
∴log2 √ab≥2
∴√ab≥4
∴ab≥16
If and only if a = b = 4, take "="

Let a > 0, b > 0, if 1 is the median of log2b and log2a, then the minimum value of 1 / A ^ 2 + 1 / b ^ 2

Given that a > 1 and a ^ (LGB) = 2 under the root of 4 times, find the minimum value of log2 (AB)

A ^ (LGB) = 2 ^ (lgblog2 (a)) = 2 ^ (1 / 4) lgblog2 (a) = log2 (b) log2 (a) / log2 (10) = 1 / 4 log2 (a) log2 (b) = log2 (10) / 4log2 (AB) = log2 (a) + log2 (b) > = 2 √ [log2 (a) log2 (b)] = 2 * (1 / 2) * √ log2 (10) = √ log2 (10) minimum √ log2 (10)