Question

Log3 (1 - x) < log3 (x + 3)

Answer 1

Запишем область определения:

$\left \{ {{1 - x > 0} \atop {x+3 > 0}} \right. \\\left \{ {{x -3}} \right.$

Оба логарифма по основанию 3, поэтому значки логарифмов мы можем откинуть:

1-х < х+3

1-3 < х+х

-2 < 2х

х > -1.

Чертим числовые оси для всех неравенств (в том числе и для неравенств из области определения), находим промежуток, который совпадает на всех числовых осях, и записываем ответ.

Ответ: ( -1;1)

Answer 2

To solve the inequality log3(1 - x) < log3(x + 3), we need to consider the domain of validity for the logarithmic functions and the properties of logarithms.

First, let's establish the domain for which the logarithmic functions are defined. The argument of a logarithm must be greater than zero:

1 - x > 0 and x + 3 > 0

Solving these inequalities individually:

1. 1 - x > 0 -x > -1 x < 1

2. x + 3 > 0 x > -3

So, the valid domain for the logarithmic functions is -3 < x < 1.

Next, we'll consider the properties of logarithms. For any positive numbers a, b, and base c (where c > 0 and c ≠ 1):

1. If a > b, then logc(a) > logc(b).

Now, let's compare the expressions log3(1 - x) and log3(x + 3):

1. log3(1 - x) < log3(x + 3)

Since the bases are the same (both are 3), we can drop the logarithm notation and work with the arguments:

1 - x < x + 3

Now, solve for x:

2x > -2 x > -1

However, this solution contradicts the domain we established earlier (-3 < x < 1). Therefore, there is no valid solution for the inequality log3(1 - x) < log3(x + 3) within the given domain.

In summary, the inequality log3(1 - x) < log3(x + 3) has no solutions in the interval -3 < x < 1.