Log2x + log4x + log8x = 11 find x

Log2x + log4x + log8x = 11 find x

Original formula = log2x + log (2) ^ 2 (x) ^ 1 + log (2 ^ 3) (x) ^ 1
=log2x+(1/2)log2x+(1/3)log2x
=(11/6)log2x=11
So log2x = 6 gets x = 64

Why can we get y = Log1 / 2x = - log2x by using the formula of changing bottom

y=log1/2x
Change to logarithm with base 2
y=[log2 x]/[log2 (1/2)]
Because log2 (1 / 2) = log2 (2 ^ - 1) = - 1
So: y = - log2x

Given - 3 ≤ Log1 / 2x ≤ - 1 / 2, how to get 1 / 2 ≤ log2x ≤ 3

∵log1/2X=﹣log2X
∴-3≤﹣log2X≤-1/2
∴1/2≤log2X≤3

There are several solutions to the equation where the square of x equals Log1 / 2x
By the way, what are they

A solution, (√ 2 / 2, √ 2 / 2)

If the minimum value of the solution of the inequality x ≥ 2 is a and the maximum value of the solution of the inequality x ≤ 5 is B, then what is (2a-b) to the power of 2010

The minimum solution of X ≥ 2 is a = 2
The maximum value of the solution of X ≤ 5 is b = 5
(2a-b) 2010th power = (4-5) 2010 times = 1

Inequality equations: x + 2 > m + n X-1

The simplified x < M-1 + 1 results in x < m; x > m + n-2
M = 2, n = negative 1
M + n to the power of 2008 = (2-1) to the power of 2008 = 1

The absolute value of X - 2 / x + 1 > 0

|x|-2/x+1>0
1. If x > 0
x - 2/x + 1 > 0
x^2 + x - 2 > 0
(x - 1)(x + 2)>0
x > 1
2. If X0
x+ 2/x - 1 >0
x < 0.

If the absolute value of inequality X / X-1 is greater than X / X-1, the solution set of inequality X / X-1 is? And there must be a specific process,

|a|>a
If a ≥ 0, then | a | = a
So a

Given that a is a nonnegative real number, the inequality AX2 - (a + 1) x + 1 about X is solved

When a = 0, the original inequality is - x + 1 < 0, and the solution set of the original inequality is (1, + ∞) (2 points) (II) when a ＞ 0, the inequality is transformed into (x-1) (AX-1) ＜ 0, and the two corresponding equations (x-1) (AX-1) = 0 are 1 and 1a. When 0 ＜ a ＜ 1, 1A ＞ 1, the original inequality solution set is (1, 1A) (4 points) when a = 1, 1A = 1, the solution set of the original inequality is (6 points) when a ＞ 1, 1A ＜ 1, the solution set of the original inequality is (1a, 1) (8 points)

Solving inequality (ax-a ^ 2-1) (X-2) ≥ 0,

(ax－a²－1)(x－2)≥0
(1) If a = 0, then the solution set of the inequality is: {x | x ≥ 2}
(2) If 0