Question

Решить неравенство Log_2(2x-3) Log_2(x+2)

Answer 1

решение

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Answer 2

To solve the inequality involving logarithms, we first need to find the domain of the expression to ensure that the logarithms are well-defined. For the given inequality:

$\log_2(2x - 3) \cdot \log_2(x + 2)$

The domain is determined by the conditions that the arguments of both logarithms must be strictly greater than zero:

$2x - 3 > 0 \quad \text{and} \quad x + 2 > 0$

Solving these inequalities separately:

$2x - 3 > 0$ $2x > 3$ $x > \frac{3}{2}$

$x + 2 > 0$ $x > -2$

So, the domain of the expression is: $x > \frac{3}{2}$.

Now, to solve the inequality:

$\log_2(2x - 3) \cdot \log_2(x + 2) > 0$

We can use the properties of logarithms:

$\log_b(a) > 0 \iff a > 1$

So, the inequality can be written as:

$\log_2(2x - 3) > 0 \quad \text{and} \quad \log_2(x + 2) > 0$

Now, solving each part separately:

$\log_2(2x - 3) > 0$

Using the logarithmic inequality rule, we have:

$2x - 3 > 1$

$2x > 4$

$x > 2$

$\log_2(x + 2) > 0$

Using the logarithmic inequality rule, we have:

$x + 2 > 1$

$x > -1$

Now, since both $\log_2(2x - 3)$ and $\log_2(x + 2)$ are positive for $x > 2$ and $x > -1$ respectively, the final solution is the intersection of these intervals, which is:

$x > 2$

So, the solution to the inequality is $x > 2$, considering the domain $x > \frac{3}{2}$.