Question

Lg(x-2)+lg(27-x)<2Pamagite​

Answer 1

Ответ:

ОДЗ:

$\left \{ {{x - 2 > 0} \atop {27 - x > 0} } \right. \\ \\ \left \{ {{x > 2} \atop {x < 27} } \right. \\ \\ x\in(2;27)$

$lg(x - 2) + lg(27 - x) < 2 \\ lg(x - 2)(27 - x) < 2 \\ (x - 2)(27 - x) < {10}^{2} \\ 27x - {x}^{2} - 54 + 2x - 100 < 0 \\ - {x}^{2} + 29x - 154 < 0 \\ {x}^{2} - 29x + 154 > 0 \\ D= 841 - 616 = 225 \\ x_1 = \frac{29 + 15}{2} = \frac{44}{2} = 22 \\ x_2 = \frac{14}{2} = 7 \\ + \: \: \: \: \: \: - \: \: \: \: \: \: \: \: + \\ - -7 - - 22 - - > \\ x\in( - \infty; 7)U(22; + \infty )$

Пересекаем с ОДЗ

Ответ:

$x\in(2;7)U(22 ;27)$

Answer 2

To solve the inequality lg(x-2) + lg(27-x) < 2Pamagite, I assume you mean that "lg" represents the logarithm with base 10 (common logarithm), and "Pamagite" is a constant value.

Here's how you can solve this inequality:

1. Start by combining the two logarithms using the properties of logarithms. The sum of logarithms with the same base can be combined into a single logarithm.

lg[(x - 2)(27 - x)] < 2Pamagite

2. Now, remove the logarithm by exponentiating both sides. Since the base is 10, you will raise 10 to the power of both sides.

10^(lg[(x - 2)(27 - x)]) < 10^(2Pamagite)

3. The left side simplifies to (x - 2)(27 - x), and on the right side, you have:

100Pamagite

So, you have:

(x - 2)(27 - x) < 100Pamagite

4. Expand the left side:

27x - x^2 - 2x + 2 < 100Pamagite

5. Rearrange and simplify the inequality:

-x^2 + 25x - 98 < 100Pamagite

6. Now, isolate x:

-x^2 + 25x - 98 - 100Pamagite < 0

7. Finally, you can solve this quadratic inequality. You can use the quadratic formula to find the roots of the quadratic equation and determine the intervals where the inequality is satisfied. Since Pamagite is a constant, it won't affect the roots.

The solutions to this inequality will depend on the value of Pamagite. You can find the roots of the quadratic equation and then determine the intervals where the expression is less than zero for the specific value of Pamagite.