Question

Log1/2(x)+log1/2(10-x)>=-1+log1/2(4,5)​

Answer 1

$\displaystyle\bf\\ODZ:\\\\\left \{ {{x > 0} \atop {10-x > 0}} \right. \ \ \Rightarrow \ \ \left \{ {{x > 0} \atop {x < 10}} \right. \ \ \ \Rightarrow \ \ \ x\in\Big(0 \ ; \ 10\Big)\\\\\\\log_{\frac{1}{2} } x+\log_{\frac{1}{2} } (10-x)\geq -1+\log_{\frac{1}{2} } 4,5\\\\\\\log_{\frac{1}{2} } x+\log_{\frac{1}{2} } (10-x)\geq \log_{\frac{1}{2} } 4,5-\log_{\frac{1}{2} } \frac{1}{2} \\\\\\\log_{\frac{1}{2} } \Big[x\cdot (10-x)\Big]\geq \log_{\frac{1}{2} } \frac{4,5}{0,5}$

$\displaystyle\bf\\\log_{\frac{1}{2} } \Big[x\cdot (10-x)\Big]\geq \log_{\frac{1}{2} } 9\\\\\\0 < \frac{1}{2} < 1 \ \ \ \Rightarrow \ \ \ x\cdot(10-x)\leq 9\\\\\\-x^{2} +10x-9\leq 0\\\\\\x^{2} -10x+9\geq 0\\\\\\(x-1)(x-9)\geq 0\\\\\\+ + + + + [1] - - - - - [9] + + + + + \\\\\\x\in\Big(-\infty \ ; \ 1\Big] \ \cup \ \Big[9 \ ; \ \infty \Big)$

С учётом ОДЗ окончательный ответ :

$\displaystyle\bf\\\boxed{x\in\Big(0 \ ; \ 1\Big] \ \cup \ \Big[9 \ ; \ 10\Big)}$

Answer 2

Ответ:

хє (0;1) U [9;10)

Пошаговое объяснение:

log½(x)+log½(10-x) ≥ -1+log½(4,5)

ОДЗ: х≤0

упрощаем выражение

log½ (x×(10-x) ≥ -1+log½(4,5)

log½(10x-x²) ≥ -1+log½(9/5)

log½(10x-x²) ≥ -1+log2-¹(4,5)

log½(10x-x²) ≥ -1-log2¹(4,5)

log½(10x-x²) ≥ -1-(log2(9)-log2(2)) (log2(2)=1)

log½(10x-x²) ≥ -1-log2(9+1)

log½(10x-x²) ≥ -log2(9)

log½(10x-x²) ≥ -log2(3²)

log½(10x-x²) ≥ -2log2(3)

при 0<а<1, выражение loga(x)≥b равносильно х≤а в степени b

10x-x²≤(½) в степени -2log2(3)

10x-x²≤ (2 в степени -1) и в степени -2log2(3)

10x-x²≤ 2 в степени 2log2(3)

10x-x²≤ 2 в степени log2(3²)

10x-x²≤ 3²

10x-x²≤ 9

10x-x²-9≤0

-x²+10x-9≤0

-x²+9x-x-9≤0

-x×(x-9)+x-9

(x-9)×(-x+1)≤0

{x-9≤0 {x-9≥0 {x≤9 {x≥9

{-x+1≥0 {-x+1≤0 {x≤1 {x≥1

xє (-∞;1] U [9;+∞), xє (0;10)

хє (0;1) U [9;10)

Answer 3

To solve the inequality, we can simplify the expression using logarithm properties and then solve for the variable. Let's start by rewriting the equation:

log₁/₂(x) + log₁/₂(10 - x) ≥ -1 + log₁/₂(4.5)

Using the property log(a) + log(b) = log(a * b), we can combine the logarithms on the left side:

log₁/₂(x * (10 - x)) ≥ -1 + log₁/₂(4.5)

Now, let's simplify further. The right side of the inequality can be rewritten as a single logarithm:

log₁/₂(x * (10 - x)) ≥ log₁/₂(4.5 / 2)

Using the property log(a) - log(b) = log(a / b), we simplify the right side:

log₁/₂(x * (10 - x)) ≥ log₁/₂(2.25)

Since the base of the logarithm is 1/2, we can rewrite the inequality as an exponentiation:

x * (10 - x) ≥ 2.25

Next, we can rearrange the equation to solve for x:

10x - x² ≥ 2.25

Rearranging the terms, we have a quadratic inequality:

x² - 10x + 2.25 ≤ 0

To solve this inequality, we can factorize or use the quadratic formula. Factoring doesn't seem straightforward in this case, so let's use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

For our equation, a = 1, b = -10, and c = 2.25:

x = (-(-10) ± √((-10)² - 4(1)(2.25))) / (2 * 1)

Simplifying further:

x = (10 ± √(100 - 9)) / 2 x = (10 ± √91) / 2

Approximately:

x ≈ (10 + √91) / 2 or x ≈ (10 - √91) / 2

Therefore, the solution to the inequality is:

(10 + √91) / 2 ≤ x ≤ (10 - √91) / 2