Question

|1-log1/6 x|+2=|3-log1/6 x| |x^2+3x-70|>=0 2|15+2x-x^2|=<-5

Answer 1

Смотрите ответ на фотографии

Answer 2

Let's solve the given equations step by step:

1. |1 - log(1/6)x| + 2 = |3 - log(1/6)x|

We'll solve this equation in two separate cases:

Case 1: (1 - log(1/6)x) + 2 = (3 - log(1/6)x) Case 2: (1 - log(1/6)x) + 2 = - (3 - log(1/6)x)

Case 1: (1 - log(1/6)x) + 2 = (3 - log(1/6)x)

Combine like terms: 3 - log(1/6)x = 3 - log(1/6)x

The equation becomes: 0 = 0

Since this is true for all x, all values of x are solutions for Case 1.

Case 2: (1 - log(1/6)x) + 2 = - (3 - log(1/6)x)

Combine like terms: 3 - log(1/6)x = -3 + log(1/6)x

Add log(1/6)x to both sides and subtract 3 from both sides: 2 log(1/6)x = -3

Divide both sides by 2: log(1/6)x = -3/2

Now, we need to solve for x. Remember that the logarithm function is only defined for positive values, so the argument (1/6)x must be greater than zero.

1/6x > 0

Divide both sides by 1/6: x > 0

Now, apply the logarithm properties:

log(1/6)x = -3/2 x = (1/6)^(-3/2) x = (6/1)^(3/2) x = 6^(3/2) x = 36

So, for Case 2, the solution is x = 36.

Therefore, the solutions for the original equation |1 - log(1/6)x| + 2 = |3 - log(1/6)x| are x > 0 and x = 36.

2. |x^2 + 3x - 70| >= 0

An absolute value is always non-negative, so the inequality |x^2 + 3x - 70| >= 0 is true for all real values of x.

3. 2|15 + 2x - x^2| < -5

The absolute value on the left side of the inequality will always be non-negative, but the right side is negative (-5), which means there are no real values of x that satisfy this inequality. In other words, there are no solutions for this inequality.

To summarize:

1. The solutions for the equation |1 - log(1/6)x| + 2 = |3 - log(1/6)x| are x > 0 and x = 36.
2. The inequality |x^2 + 3x - 70| >= 0 is true for all real values of x.
3. There are no solutions for the inequality 2|15 + 2x - x^2| < -5.