Question

Log2(2x^2+4)-log2(x^2-x+10)>=log2(2-1/x)

Answer 1

$\displaystyle log_2(2x^2+4)-log_2(x^2-x+10) \geq log_2(2- \frac{1}{x})\\\\ODZ: \\\\ 2x^2+4\ \textgreater \ 0; x\in R\\\\x^2-x+10\ \textgreater \ 0; x\in R\\\\ \frac{2x-1}{x}\ \textgreater \ 0; x\in (-oo;0)(1/2;+oo)$

решение: 
$\displaystyle log_2 \frac{2x^2+4}{x^2-x+10} \geq log_2( \frac{2x-1}{x})\\\\ \frac{(2x^2+4)*x-(2x-1)(x^2-x+10}{x(x^2-x+10)} \geq 0\\\\ \frac{2x^3+4x-2x^3+3x^2-21x+10}{x(x^2-x+10)} \geq 0\\\\ \frac{3x^2-17x+10}{x(x^2-x+10)} \geq 0\\\\ \frac{(x-5)(3x-2)}{x(x^2-x+10)} \geq 0$


____-___ 0 ____+_____2/3_____-___5____+___

(0;2/3][5;+oo)

с учетом ОДЗ
ОТВЕТ (1/2; 2/3][5;+oo)

Answer 2

To solve the inequality log₂(2x² + 4) - log₂(x² - x + 10) ≥ log₂(2 - 1/x), we can use properties of logarithms to simplify and solve it.

First, remember that you can combine logarithms with subtraction inside a single logarithm using the division property of logarithms:

log₂(a) - log₂(b) = log₂(a/b)

So, the original inequality becomes:

log₂((2x² + 4) / (x² - x + 10)) ≥ log₂(2 - 1/x)

Now, we have an inequality of logarithms with the same base (log base 2), which means the arguments inside the logarithms must be equal:

(2x² + 4) / (x² - x + 10) ≥ (2 - 1/x)

Now, we need to solve this inequality. To make the calculations more manageable, let's get rid of the logarithms and work with the inequalities directly:

(2x² + 4) / (x² - x + 10) ≥ (2 - 1/x)

Now, we'll get rid of the fraction by multiplying both sides of the inequality by the least common denominator, which is x(x² - x + 10) to clear the fractions:

x(x² - x + 10) * [(2x² + 4) / (x² - x + 10)] ≥ x(x² - x + 10) * [(2 - 1/x)]

This simplifies to:

2x(x² + 2) ≥ x(x² - x + 10)(2 - 1/x)

Now, distribute on both sides:

2x³ + 4x ≥ 2x(x² - x + 10) - (x³ - x² + 10x)

Simplify further:

2x³ + 4x ≥ 2x³ - 2x² + 20x - x³ + x² - 10x

Now, combine like terms:

2x³ + 4x ≥ 2x³ - x³ - 2x² + x² + 20x - 10x

Simplify:

2x³ + 4x ≥ x³ - x² + 10x

Now, subtract x³ from both sides to isolate the x² term:

2x³ - x³ + 4x - x³ ≥ -x² + 10x

This simplifies to:

x³ + 4x - x³ ≥ -x² + 10x

The x³ terms cancel out:

4x ≥ -x² + 10x

Now, subtract 10x from both sides:

4x - 10x ≥ -x²

Simplify:

-6x ≥ -x²

Now, divide both sides by -1, and remember to reverse the inequality since you're dividing by a negative number:

6x ≤ x²

Now, subtract 6x from both sides:

0 ≤ x² - 6x

Factor the right side:

0 ≤ x(x - 6)

Now, you have two factors to consider:

1. x ≥ 0
2. x - 6 ≤ 0, which implies x ≤ 6

So, the solution to the original inequality is:

0 ≤ x ≤ 6

This is the range of values for x that satisfies the inequality log₂(2x² + 4) - log₂(x² - x + 10) ≥ log₂(2 - 1/x).