Let f (x) = ACOS ^ 2 (Wx) - √ 3asinwxcoswx + B have the minimum positive period π (1) When a = - 2, B = 1, find the symmetry axis equation and symmetry center of F (x) image (2) If the domain of F (x) is 〔 - π / 3, π / 6 〕, and the range of F (x) is 〔 - 1,5 〕, the values of a and B are obtained

Let f (x) = ACOS ^ 2 (Wx) - √ 3asinwxcoswx + B have the minimum positive period π (1) When a = - 2, B = 1, find the symmetry axis equation and symmetry center of F (x) image (2) If the domain of F (x) is 〔 - π / 3, π / 6 〕, and the range of F (x) is 〔 - 1,5 〕, the values of a and B are obtained

(1) F (x) = ACOS ^ 2 (Wx) - √ 3asinwxcoswx + B = a (1 + cos2wx) / 2-radical 3asin2wx / 2 + B = A / 2 + acos2wx / 2-radical 3asin2wx / 2 + B = asin (PI / 6-2wx) + A / 2 + B period T = 2pi / 2W = Pi, w = 1, so f (x) = asin (PI / 6-2x) + A / 2 + B, so symmetry axis X = - KPI / 2-pi / 6 symmetry center (- KP

If the square of the absolute value of M-1 of (M + 2) x + 5Y - 2 = 0 is a quadratic equation with respect to X. y, try to find the value of M?

(m+2)x+5y^(|m-1|^2)-2=0
|M-1 | ^ 2 = 1 and M + 2 is not equal to 0
M = 0 or M = 2

Given that the square of the quadratic equation (m-2) x + 3x + M-4 = 0, there is a root of 0, find M

Let x = 0
Equation with people
m²-4=0
m=±2
When m = 2
The equation is not a quadratic equation of one variable
So m = - 2

Log3 5 = a log3 7 = B log25 7
I don't understand how it's deformed here

Bottom changing formula
a=lg5/lg3
lg5=alg3
b=lg7/lg3
lg7=blg3
So the original formula = lg7 / LG25
=lg7/2lg5
=blg3/2alg3
=b/(2a)

If a = log3 (10), B = log3 (7), then 3 ^ A-B =?

3^(a-b)
=3^(log3(10)-log3(7))
=3^[log3(10/7)]
=10/7

Simplification (log3 4) / (log9 8)

If we take nine as the square of three, then the denominator can be taken as 1 / 2 log3 8
(Note: 1 / 2 here has not been divided into molecules) then the whole formula can be resolved to 2log3 4 / log3 8
Next, if you don't know, you can ask me a question directly. I'll use the computer to help you answer later

Why log9 (2) = log3 (2) / 2
Such as the title

The formula of changing bottom: loga (b) = logC (b) / logC (a)
On the left, change to base 3
=Log3 (2) / log3 (9) = right

It is known that M = log36,3 to the nth power = 5, and m and N are used to denote log5 root sign 18

M = log3 (6) n = log3 (5) using the bottom changing formula log5 (√ 18) = (1 / 2) log5 (18) = (1 / 2) [log3 (18) / log3 (5)] = (1 / 2) [log3 (3 * 6) / log3 (5)] = (1 / 2) [log3 (3) + log3 (6)] / log3 (5) = (1 / 2) (1 + m) / N = (1 + m) / (2n)

Given 5 ^ a = 4, log5 3 = B, prove log125 12 = 1 / 3 (a + b)
Attached problem solving process, thank you!

5^a=4,log5 3=b
==>log5 4=a
log125 12=log(5^3) 12
=(1/3)log5 12 =(1/3)log5 3*4
=1/3)(log5 3 +log5 4)
=(1/3)(a+b)

A = [log5 (4)] square B = [log5 (3)] compare the size of a and B! How to prove?
The question was originally one to choose, the correct answer is "a big" but I choose the wrong, I think "B big" to solve the problem method!

Proof thinking
Firstly, the logarithmic function with base 5 is monotone increasing function in its domain of definition,
4 > 3,
The logarithm of base 4 with 5 > the logarithm of base 3 with 5,
And then
Logarithm of base 4 with 5 > logarithm of base 3 with 5 > logarithm of base 1 with 5 = 0,
Get
[logarithm of base 4 with 5] &# 178; > [logarithm of base 3 with 5] &# 178; > logarithm of base 3 with 5
So the correct answer is "a big"