Question

Log2,5(6-x)>=log2,5(4-3x)​

Answer 1

***

$log_{2,5}(6 - x) \geq log_{2,5} (4 - 3x)$

$6-x > 0$

$4 - 3x > 0$     =>    неравенство определяет множество  (–∞; 4/3)

Поскольку основание 2,5 > 1

то используя свойства логарифмов:

$6 - x \geq 4 - 3x$

$-x+3x \geq 4-6$

$2x \geq -2$

$x \geq \frac{-2}{2}$

$x\geq -1$       =>   неравенство определяет множество [-1; +∞)

(–∞; 4/3) ∩ [-1; +∞) = [-1; 4/3)

х ∈ [-1; 4/3)

ответ: [-1; 4/3)

Answer 2

$\displaystyle\bf\\ODZ :\\\\\left \{ {{6-x > 0} \atop {4-3x > 0}} \right. \ \ \ \Rightarrow \ \ \ \left \{ {{x < 6} \atop {x < 1\dfrac{1}{3} }} \right. \ \ \ x < 1\frac{1}{3} \\\\\\\log_{2,5} \Big(6-x\Big)\geq \log\Big(4-3x\Big)\\\\\\2,5 > 1 \ \ \ \Rightarrow \ \ \ 6-x\geq 4-3x\\\\\\-x+3x\geq 4-6\\\\\\2x\geq -2\\\\\\x\geq -1\\\\\\Otvet \ : \ x\in\Big[-1 \ ; \ 1\frac{1}{3} \Big)$

Answer 3

To solve the inequality log₂.₅(6 - x) ≥ log₂.₅(4 - 3x), we can follow these steps:

Step 1: Convert both sides of the inequality to exponential form using the base 2.5.

2.5^(log₂.₅(6 - x)) ≥ 2.5^(log₂.₅(4 - 3x))

Step 2: Simplify using the property that exponentiation is the inverse of logarithms.

6 - x ≥ 4 - 3x

Step 3: Solve the resulting linear inequality.

2x ≥ 2

Step 4: Divide both sides by 2 to isolate x.

x ≥ 1

Therefore, the solution to the inequality log₂.₅(6 - x) ≥ log₂.₅(4 - 3x) is x ≥ 1.