Question

Log (0,7) 2x-1/x-1 >=0

Answer 1

Решение на фотографии

Answer 2

To solve the inequality $\log_{0.7}\left(\frac{2x-1}{x-1}\right) \geq 0$, we need to find the values of $x$ that satisfy this inequality.

Step 1: Domain Restriction First, note that the logarithm function $\log_{0.7}(y)$ is only defined when $0.7 > 0$ and $y > 0$ since the base of the logarithm (0.7 in this case) must be positive, and the argument of the logarithm (in this case, $\frac{2x-1}{x-1}$) must be positive. Therefore, we need to find the values of $x$ that make the argument of the logarithm positive.

Step 2: Determine the Sign of the Argument of the Logarithm To find when the argument of the logarithm is positive, set the argument greater than zero and solve for $x$:

$\frac{2x-1}{x-1} > 0$

Step 3: Find Critical Points To find critical points, set the numerator and denominator equal to zero and solve for $x$:

$2x - 1 = 0 \Rightarrow x = \frac{1}{2}$

$x - 1 = 0 \Rightarrow x = 1$

Step 4: Sign Analysis of the Inequality Now, let's perform a sign analysis on the inequality $\frac{2x-1}{x-1} > 0$ by considering different intervals of $x$:

Interval 1: $x < \frac{1}{2}$ Choose a test point $x_1 < \frac{1}{2}$ (e.g., $x_1 = 0$), then $\frac{2x_1-1}{x_1-1} = \frac{-1}{-1} = 1 > 0$

Interval 2: $\frac{1}{2} < x < 1$ Choose a test point $x_2$ such that $\frac{1}{2} < x_2 < 1$ (e.g., $x_2 = \frac{3}{4}$), then $\frac{2x_2-1}{x_2-1} = \frac{1}{\frac{1}{4}} = 4 > 0$

Interval 3: $x > 1$ Choose a test point $x_3 > 1$ (e.g., $x_3 = 2$), then $\frac{2x_3-1}{x_3-1} = \frac{3}{1} = 3 > 0$

Step 5: Combine Intervals We found that $\frac{2x-1}{x-1} > 0$ for $x < \frac{1}{2}$ and $\frac{1}{2} < x < 1$ and $x > 1$.

Step 6: Final Solution Since the inequality is $\log_{0.7}\left(\frac{2x-1}{x-1}\right) \geq 0$, the solution is all values of $x$ in the combined intervals where the argument of the logarithm is positive:

$\frac{2x-1}{x-1} > 0$ for $x < \frac{1}{2}$ and $\frac{1}{2} < x < 1$