Question

Log_4(5x + 1) < log_4(3 - 2x)

Answer 1

Ответ: x∈(-0.2; 2/7)

Объяснение:

ОДЗ:  5x+1>0 => x> -1/5     3-2x>0 => x<1.5

Итак ОДЗ   x∈ (-0.2 ; 1.5)

Решаем неравенство

Так как основания логарифмов одинаковые и больше 1, то можно заменить данное неравенство на эквивалентное

5x+1<3-2x

7x<2

x<2/7

C учетом ОДЗ  x∈(-0.2; 2/7)

Answer 2

To solve the inequality log₄(5x + 1) < log₄(3 - 2x), you can use the properties of logarithms.

The key property to remember is that if you have logₐ(b) < logₐ(c), where a > 0 and a ≠ 1, then it implies that b < c.

In your case, you have:

log₄(5x + 1) < log₄(3 - 2x)

Now, using the property mentioned above:

5x + 1 < 3 - 2x

Let's isolate the x terms on one side:

5x + 2x < 3 - 1

7x < 2

Now, divide both sides by 7 (since 7 is positive, you don't need to flip the inequality sign):

x < 2/7

So, the solution to the inequality log₄(5x + 1) < log₄(3 - 2x) is:

x < 2/7