Question

Log0.2 (x^2-x-2)>log 1/5 (-x^2+2x+3)

Answer 1

Ответ (2;2,5)

Решение неравенства

Корни находите удобным вам способом.

Answer 2

Ответ:

x принадлежит промежутку (2; 2,5)

Answer 3

To solve the inequality $\log_{0.2}(x^2 - x - 2) > \log_{1/5}(-x^2 + 2x + 3)$, we can use the properties of logarithms to simplify and solve for $x$. Here's the step-by-step process:

Step 1: Remove the logarithms using their properties. $\log_{0.2}(x^2 - x - 2) > \log_{1/5}(-x^2 + 2x + 3)$

Step 2: Apply the logarithmic property $\log_a(b) > \log_a(c)$ implies $b > c$. $x^2 - x - 2 > -x^2 + 2x + 3$

Step 3: Combine like terms on both sides of the inequality. $2x^2 - 3x - 5 > 0$

Step 4: Solve the quadratic inequality. You can do this by factoring or using the quadratic formula. In this case, let's use factoring.

$2x^2 - 3x - 5 > 0$ can be factored as $(2x + 1)(x - 5) > 0$

Now, we have two critical points: $x = -\frac{1}{2}$ and $x = 5$.

Step 5: Create intervals based on the critical points and test points within each interval to determine the sign of the expression $2x^2 - 3x - 5$.

Test $x = 0$: $2(0)^2 - 3(0) - 5 = -5 < 0$ Test $x = 2$: $2(2)^2 - 3(2) - 5 = 1 > 0$ Test $x = 6$: $2(6)^2 - 3(6) - 5 = 37 > 0$

So, the solution to the inequality $2x^2 - 3x - 5 > 0$ is $x < -\frac{1}{2}$ or $x > 5$.

Finally, this solution needs to be checked in the original inequality to ensure it is valid since logarithms only accept positive arguments.

$\log_{0.2}(x^2 - x - 2) > \log_{1/5}(-x^2 + 2x + 3)$

Check $x < -\frac{1}{2}$: In this interval, $x^2 - x - 2$ is positive, and $-x^2 + 2x + 3$ is also positive. So, the inequality holds.

Check $x > 5$: In this interval, $x^2 - x - 2$ is positive, and $-x^2 + 2x + 3$ is negative. Thus, the inequality does not hold.

Therefore, the valid solution is $x < -\frac{1}{2}$.