Question

Log1/3 (2x-1/x+2)> 1

Answer 1

$\log_{\frac{1}{3}}(\dfrac{2x-1}{x+2}) > 1 \\ \\ ODZ: \ \left\{ \begin{gathered} \dfrac{2x-1}{x+2}>0 \\ x + 2\ne 0 \\ \end{gathered} \right. \ ; \ \left\{ \begin{gathered} \dfrac{2(x-\dfrac{1}{2})}{x+2} > 0 \\ x \ne - 2 \\ \end{gathered} \right. \ ; \ x \in (-\infty ; -2)\cup (\dfrac{1}{2}; +\infty)$

$\log_{\frac{1}{3}}(\dfrac{2x-1}{x+2}) >\log_{\frac{1}{3}}(\dfrac{1}{3})^{1} \\ \\ \dfrac{2x-1}{x+2} < \dfrac{1}{3} \\ \\ \dfrac{2x-1}{x+2} - \dfrac{1}{3}< 0 \\ \\ \dfrac{3(2x-1) - 1(x+2)}{3(x+2)} < 0 \\ \\ \dfrac{5x-5}{3(x+2)} < 0 \\ \\ \dfrac{5(x-1)}{3(x+2)} < 0 \ / * \dfrac{3}{5} \\ \\ x \in (-2;1)$

С учётом ОДЗ (1):

$x \in (\dfrac{1}{2} ; 1)$

Ответ: x ∈ (1/2 ; 1)

Примечание: если основание логарифма относится к промежутку от 0 до 1, то знак неравенства меняется в противоположную сторону.

Answer 2

To solve the inequality log₁/₃(2x - 1/x + 2) > 1, we need to be careful about the logarithm's base. It seems like you intended to use a base of 1/3 for the logarithm. Please confirm if this is correct, as the parentheses are not very clear. I'll assume the inequality is:

log₁/₃(2x - 1/x + 2) > 1

To solve this, we'll follow these steps:

Step 1: Get rid of the logarithm by expressing the inequality as an exponential equation. Step 2: Solve the exponential equation for x. Step 3: Determine the valid interval(s) for x that satisfy the inequality.

Step 1: Exponential form To get rid of the logarithm with base 1/3, we'll rewrite the inequality using exponential notation:

1. 1/3 raised to what power equals (2x - 1/x + 2)?

The exponential form is:

1/3^(power) > 2x - 1/x + 2

Step 2: Solve for x Now, we need to solve the exponential equation for x.

1/3^(power) > 2x - 1/x + 2

Step 3: Determine the valid interval(s) for x Without knowing the exact expression in the parentheses of the logarithm, we cannot find an exact solution. However, we can make some general observations based on the properties of logarithms.

In the given inequality, we have a logarithm of base 1/3, which means the argument inside the logarithm must be greater than 1 for the logarithm to exist. So, we need to ensure:

2x - 1/x + 2 > 1

Now, the above inequality is quadratic in nature due to the presence of x and 1/x terms. To solve this, we'll combine the x terms:

(2x² - 1 + 2x) / x > 1

Multiply both sides by x to clear the fraction:

2x² - 1 + 2x > x

Now, bring all the terms to one side:

2x² - x + 1 > 0

We don't have an exact solution yet, but the quadratic term suggests this is a parabola that opens upwards. The inequality should hold for specific ranges of x. The exact range will depend on the expression inside the logarithm, which is missing in your question.

Without additional information, I cannot find the exact solution or provide an interval for x that satisfies the inequality. If you can provide the complete expression inside the logarithm, I'd be able to help you find a more precise solution.