If inequality LG (2aX) / LG (a + x)

If inequality LG (2aX) / LG (a + x)

lg(2ax)/lg(a+x)

When 1 ＜ x ≤ 2, the inequality x2-2ax + a ＜ 0 holds and the value range of real number a is obtained

Let f (x) = x2-2ax + A, then from the image and properties of quadratic function, when 1 < x ≤ 2, the inequality x2-2ax + a < 0 holds, which is equivalent to, f (1) ≤ 0, f (2) < 0, that is, 1 − 2A + a ≤ 04 − 4A + a < 0, the solution, a ＞ 43, the value range of real number a (43, + ∞)

When a ＞ 0, and a is not equal to 1, the inequality a ^ [(2 ^ x) + x)] < a [(x ^ 2) + 3x + 3]
The wrong topic is a ^ [(2 times x ^ 2) + x)] < a [(x ^ 2) + 3x + 3]

00
(x-3)(x+1)>0
x3
a>1
A ^ x is an increasing function
So 2x ^ 2 + X

Given the function f (x) = ax ^ 2 + BX + C (a is not equal to 0), a, B. C belong to R, set a =, when a has only 2, a: C

∵f（x）=x,
∴ax^2+bx+c=x
ax^2+（b-1）x+c=0
∵ set a = {f (x) = x}, a has only 2 elements
∴（b-1）^2-4ac=0①
∵ a has only two elements
The parabola and the straight line y = x have and only have one intersection point
Y = x is the tangent of the parabola
∴f’（2）=0
∵f’（x）=2ax+b
∴4a+b=0
∴b=-4a②
By substituting formula (2) into formula (1), we get (- 4a-1) ^ 2-4ac = 0 and simplify it to 16A & sup2; + 8A + 1-4ac = 0 (3)
4a+2b+c=2,4a-8a+c=2,c-4a=2④
③ 4. A = - 1 / 4, C = 1
∴a/c=-1/4