Question

Log3+x 3 + log3+x x² < log3+x (x+4)

Answer 1

Ответ:

Объяснение:

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Answer 2

To solve the inequality $\log_{3+x}(3) + \log_{3+x}(x^2) < \log_{3+x}(x+4)$, we can use the properties of logarithms to simplify it.

First, let's use the fact that $\log_a(b) + \log_a(c) = \log_a(b * c)$:

$\log_{3+x}(3x^2) < \log_{3+x}(x+4)$

Now, we can eliminate the logarithms by raising the base $3+x$ to the power of both sides:

$3x^2 < x+4$

Next, let's bring all terms to one side of the inequality:

$3x^2 - x - 4 < 0$

Now, we have a quadratic inequality. To solve it, we can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = 3$, $b = -1$, and $c = -4$. Plugging these values into the quadratic formula:

$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-4)}}{2(3)}$

Simplifying further:

$x = \frac{1 \pm \sqrt{1 + 48}}{6}$

$x = \frac{1 \pm \sqrt{49}}{6}$

$x = \frac{1 \pm 7}{6}$

So, we have two possible solutions:

1. $x = \frac{1 + 7}{6} = \frac{8}{6} = \frac{4}{3}$
2. $x = \frac{1 - 7}{6} = \frac{-6}{6} = -1$

Now, we need to check the intervals where the inequality is satisfied. We can do this by testing a value in each interval.

1. Test $x = 0$: $3(0)^2 - 0 - 4 = -4$, which is less than 0.
2. Test $x = 1$: $3(1)^2 - 1 - 4 = -2$, which is less than 0.
3. Test $x = 2$: $3(2)^2 - 2 - 4 = 10$, which is greater than 0.

So, the inequality is satisfied for $x < -1$ and $-1 < x < \frac{4}{3}$. Therefore, the solution to the inequality is:

$-1 < x < \frac{4}{3}$