Question

Log₇ (4x – 6) > log₇ (2x – 4)

Answer 1

Ответ:

Пошаговое объяснение:

ОДЗ:

4x-6>0                   x>1,5

2x-4>0                    x>2

Решение:

4x-6>2x-4

2x>2

x>1

пересекаем с ОДЗ и получаем ответ x∈(2;+∞)

Answer 2

$\log_{7} (4x - 6) > \log_{7} (2x - 4)$

Данное неравенство равносильно системе:

$\left\{\begin{array}{ccc}4x - 6 > 0 \ \ \ \ \ \ \ \\2x - 4 > 0 \ \ \ \ \ \ \ \\4x - 6 > 2x - 4\end{array}\right$

$\left\{\begin{array}{ccc}x > \dfrac{3}{2} \\x > 2\\x > 1\end{array}\right$

Общим решением будет интервал $x \in (2; +\infty)$

Ответ: $x \in (2; +\infty)$

Answer 3

To solve the inequality log₇(4x - 6) > log₇(2x - 4), we'll start by applying the properties of logarithms. Remember that if you have logₐ(b) > logₐ(c), then b > c, as long as a > 1.

Given: log₇(4x - 6) > log₇(2x - 4)

Step 1: Apply the property that if two logarithms with the same base are being compared, then the inequality is determined by the values inside the logarithms.

4x - 6 > 2x - 4

Step 2: Subtract 2x from both sides to isolate the x terms on one side.

4x - 2x - 6 > -4

Step 3: Simplify the left side.

2x - 6 > -4

Step 4: Add 6 to both sides.

2x > 2

Step 5: Divide both sides by 2.

x > 1

So, the solution to the inequality is x > 1. This means any value of x that is greater than 1 will satisfy the original inequality log₇(4x - 6) > log₇(2x - 4).