Question

In(4x-3) +In1≤2In3 пожалуйста помогите с примером

Answer 1

Ответ:   $x\in (\ 0,75\ ;\ 3\ ]$

$ln(4x-3)+ln1\leq 2ln3\ \ ,\ \ \ \ ODZ:\ 4x-3 > 0\ ,\ \ x > 0,75\\\\ln(4x-3)+0\leq ln3^2\\\\ln(4x-3)\leq ln9$

Так как основание логарифма - число, большее 1, то функция   возрастающая и тогда записываем тот же знак между аргументами логарифмов:

$4x-3\leq 9\ \ \ \to \ \ \ 4x\leq 12\ \ ,\ \ x\leq 3\\\\x\in (-\infty ;\ 3\ ]$

Учитывая  ОДЗ запишем ответ:   $x\in (\ 0,75\ ;\ 3\ ]$  .

Answer 2

Ответ:

$ln(4x-3) +ln1 \leq 2ln3\\ln(4x-3)*1\leq ln3^2\\ln(4x-3)\leq ln9\\4x-3\leq 9\\4x\leq 12\\x\leq 3$

найдем ОДЗ

$4x-3 > 0\\4x > 3\\x > 0.75$

ответ

x ∈ (0.75; 3]

Answer 3

Solving the Equation In(4x-3) = In1 ≤ 2In3

To solve the equation ln(4x-3) = ln1 ≤ 2ln3, we can start by understanding the properties of natural logarithms.

Properties of Natural Logarithms: 1. ln(a) = ln(b) if and only if a = b 2. ln(a) ≤ ln(b) if and only if a ≤ b

Now, let's solve the given equation step by step.

Step 1: Simplify the Equation

The given equation ln(4x-3) = ln1 ≤ 2ln3 can be simplified using the properties of natural logarithms.

Step 2: Applying the Properties of Natural Logarithms

We know that ln1 = 0, so the equation ln(4x-3) = ln1 simplifies to ln(4x-3) = 0.

Step 3: Solving for x

Since ln(4x-3) = 0, we can apply the property that ln(a) = 0 if and only if a = 1. Therefore, 4x-3 = 1.

Step 4: Final Calculation

Solving for x: 4x - 3 = 1 4x = 4 x = 1

Conclusion

The solution to the equation ln(4x-3) = ln1 ≤ 2ln3 is x = 1.

I hope this helps! If you have any further questions or need additional assistance, feel free to ask.