The solution set of log (x + 1) base [(2x ^ 2) - 2x + 1] = 2 is

The solution set of log (x + 1) base [(2x ^ 2) - 2x + 1] = 2 is

Log (x + 1) bottom [(2x ^ 2) - 2x + 1] = log (x + 1) (x + 1) & #178;
∴2x²-2x+1=x²+2x+1
∴x=0
But when x = 0, x + 1 = 1
Base cannot be 1
The equation has no solution

If log is based on 2 and X is less than 1, then the value range of X is 0

Log is based on 2, x < 1 = log (2) 2
That is 0 < x < 2

log(1/2) |x-1|>0

A:
log1/2 |x-1| >0
The results are as follows:
0

If real numbers a and B satisfy a + 3b-2 = 0, what is the minimum value of 3 ^ A + 27 ^ B + 3?

a+3b-2=0,
Then a + 3B = 2
3^a+27^b=3^a+3^3b≥2√3^(a+3b)=2√3^2=6
So 3 ^ A + 27 ^ B + 3 ≥ 9
Take the equal sign when a = 3B

Given that real numbers x and y satisfy the conditions XY ≤ 64, X ≥ 2 and Y ≥ 2, then the minimum value of Z = log2 (x) + log2 (y) is

z=log2 (x)+ log2(y)=log2(xy)>=log2(4)=2
The minimum is 2

Given a > 0, x + y + Z = 0, XYZ is a real number, find the minimum value of log2 (1 + A ^ x) + log2 (1 + A ^ y) + log2 (1 + A ^ z)

log2（1+a^x）+log2（1+a^y）+log2（1+a^z）
=log2[1+a^y+a^x+a^(x+y)](1+a^z)
=log2[1+a^x+a^y+a^z+a^(x+y)+a^(x+z)+a^(y+z)+a^(x+y+z)]
=log2[1+a^x+a^y+a^z+a^(x+y)+a^(x+z)+a^(y+z)+1]
because
a^x+a^y+a^z+a^(x+y)+a^(x+z)+a^(y+z)≥3+3=6
The condition of equal sign is a ^ x = a ^ y = a ^ Z, that is, x = y = z = 0
So the minimum is 3

Let m = a + [1 / A-2] (2)

A & sup2; - 6A + 9 > 0 means a & sup2; - 2A + 1 > 4a-8 (A & sup2; - 2A + 1) / (A-2) > 4 means a + [1 / A-2] > 4 Log1 / 2 (x ^ 2 + 1 / 16) ≤ Log1 / 2 (1 / 16) = 4 m > n

1) Let m = a + 1 / A-2 (2)

M>N

Let m = {x | 2 ^ (x-1)

2^(x-1)＜1=2^0
∵2＞1
Ψ y = 2 ^ x is a monotonically increasing function
∴x-1＜0
∴x＜1
log1/2 x

Known Log1 / 2m

m>n>1