It is known that the function f (x) = ACOS (Wx + φ) (a > 0, w > 0, - π / 2 ≤ φ ≤ π / 2) defined on R, and the difference between the maximum and the minimum is 4 The distance between the two lowest points is π, and all the symmetry centers of the function y = sin (2x + π / 3) image are on the symmetry axis of the image y = f (x) (1) The expression of finding f (x) (2) If f (X. / 2) = 3 / 2 (x ∈ [- π / 2, π / 2], find the value of COS (X. - π / 3) (3) Let vector a = (f (x - π / 6), 1), vector b = (1, mcosx), X ∈ (0. π / 2), if vector a * vector B + 3 ≥. Constant holds, find the value range of real number M

It is known that the function f (x) = ACOS (Wx + φ) (a > 0, w > 0, - π / 2 ≤ φ ≤ π / 2) defined on R, and the difference between the maximum and the minimum is 4 The distance between the two lowest points is π, and all the symmetry centers of the function y = sin (2x + π / 3) image are on the symmetry axis of the image y = f (x) (1) The expression of finding f (x) (2) If f (X. / 2) = 3 / 2 (x ∈ [- π / 2, π / 2], find the value of COS (X. - π / 3) (3) Let vector a = (f (x - π / 6), 1), vector b = (1, mcosx), X ∈ (0. π / 2), if vector a * vector B + 3 ≥. Constant holds, find the value range of real number M

A=2；
w=2；
φ=(1/3)π;
f(x)=2cos(2x+ π/3 );
(2) f（x./2）= 2cos(x.+ π/3) =3/2
Cos (X. - π / 3) = - (3 / 8 + T) or - (3 / 8 - t) where t = 21 under the root of 8;
(3) The title is not clear

Given that f (x) = ACOS (ω x + φ) is shown in the figure, f (π 2) = - 23, then f (0)=______ ．

The minimum positive period is 2 π 3. So f (0) = f (2 π 3). Note that 2 π 3 and π 2 are symmetrical about 7 π 12, so f (2 π 3) = - f (π 2) = 23. So the answer is: 23

1. There are two propositions: 1) the equation 9 ^ x + (4 + a) × 3 ^ x + 4 = 0 about X has a solution; 2) the function f (x) = log (2a ^ 2-A) (x) is a decreasing function
There are two propositions: 1) the equation 9 ^ x + (4 + a) × 3 ^ x + 4 = 0 about X has a solution; 2) the function f (x) = log (2a ^ 2-A) (x) is a decreasing function. When there is at least one true proposition between (1) and (2), the value range of the real number a should be calculated in detail. Thank you

① It can be regarded as y = 3 ^ x, (Y > 0). Equation = > y ^ 2 + (4 + a) y + 4 = 0, using the solvable discriminant and only positive solutions (△ > = 0; Y1 * Y2 > 0; Y1 + Y2 > 0;)
Solution: {a > = 0 or a

Solving the equation: log x + 2 * log x + log x = 7

log2x+2log4x+log8x=7
Because 2log4x = 2 * 1 / 2 log2x = log2x
log8x=1/3 log2x
So log2x + 2log4x + log8x = log2x + log2x + 1 / 3 log2x = 7 / 3 log2x = 7
That is log2x = 3
So x = 8

The square of equation (1 / 3) - x-3 = log 27 with base 3

The square of (1 / 3) is x + 3 + log, and the base of 3 is 27 = x + 6
X squared - 3x-18 = 0
(x-6)(x+3)=0
X = 6 or x = - 3

The logarithm of the equation y = log with base 2 is equal to the number of solutions of π

Log is based on 2, and the logarithm of X is equal to π
π power of x = 2

The number of solutions of the equation x ^ 2 = log (1 / 2) x is

The solution makes the image of y = x ^ 2 and y = log (1 / 2) (x) (x > 0)
We know that there is only one intersection point in the image of two functions
so
The number of solutions of the equation x ^ 2 = log (1 / 2) x is 1

Log9 ^ 5 = a, log3 ^ 7 = B, find log35 ^ 9 =?

log9^5=a,log3^7=b
LOG(3,7)=2LOG(9,7)=B
==>LOG(9,7)=B/2
lOG(35,9)=1/LOG(9,35)=1/(LOG(9,7)+LOG(9,5))=1/(B/2+A)=2/(B+2A)

If log9 5 = a, log3 7 = B, then log35 9=______ .

2/(2a+b)
Because log3 7 = B, log9 7 = B / 2
So log 9 35 = log 9 (5 * 7) = log 9 5 + log 9 7 = a + B / 2
So log359 = 1 / (log935) = 1 / (a + B / 2) = 2 / (2a + b)

If log3 (4) = a, log3 (5) = B, then log9 (16 / 25)=

log9(16/25)=log9[25]-log9[16]=log3[5]-log3[4]=b-a