Question

Решить геравенство: 3^x+3^(|x|-1)≥4

Answer 1

Готово! Ответ на фото:

Answer 2

To solve the inequality $3^x + 3^{|x|-1} \geq 4$, we need to consider two cases: $x \geq 0$ and $x < 0$, since the absolute value $|x|$ makes the equation slightly more complex.

Case 1: $x \geq 0$ For $x \geq 0$, $|x| = x$, so the inequality becomes: $3^x + 3^{x-1} \geq 4$

Now, we can simplify the inequality further: $3^x + \frac{3^x}{3} \geq 4$

Combining the terms: $\frac{4}{3} \cdot 3^x \geq 4$

Dividing both sides by $\frac{4}{3}$: $3^x \geq 3$

Now, take the logarithm base 3 of both sides to isolate $x$: $x \geq \log_3(3)$

Since $\log_3(3) = 1$, the solution for this case is $x \geq 1$.

Case 2: $x < 0$ For $x < 0$, $|x| = -x$, so the inequality becomes: $3^x + 3^{-x-1} \geq 4$

Now, we can simplify the inequality further: $3^x + \frac{1}{3} \cdot 3^{-x} \geq 4$

Combining the terms: $3^x + \frac{1}{3^x} \geq 4$

To solve this inequality, let's introduce a new variable. Let $t = 3^x$, so the inequality becomes: $t + \frac{1}{t} \geq 4$

Now, we have a quadratic-like inequality. To solve it, we can multiply the whole inequality by $t$ (since $t > 0$ for $x < 0$): $t^2 + 1 \geq 4t$

Moving all terms to one side: $t^2 - 4t + 1 \geq 0$

This is a quadratic inequality, and to find the intervals where it is true, we can use the quadratic formula: $t = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$ $t = \frac{4 \pm \sqrt{12}}{2}$ $t = 2 \pm \sqrt{3}$

Since $t = 3^x$ and $x < 0$, we know that $t$ must be between 0 and 1. Therefore, the valid solution for this case is: $2 - \sqrt{3} \leq 3^x < 2 + \sqrt{3}$

Now, let's convert this back into the range of $x$ values. Take the logarithm base 3 of both sides: $x \leq \log_3(2 + \sqrt{3}) \approx 0.681$

Thus, the solution for this case is $x < 0.681$.

Final solution: Combining the solutions from both cases, we have: $x \geq 1 \quad \text{or} \quad x < 0.681$

So, the inequality is satisfied for all $x$ values greater than or equal to 1 and for $x$ values less than 0.681.