Question

2 log_2 x <2+log_2 (x+3)

Answer 1

В приложении ........

Answer 2

To solve the inequality $2\log_2 x < 2 + \log_2 (x + 3)$, we need to isolate the variable $x$ on one side of the inequality. Here's the step-by-step solution:

Step 1: Combine the logarithms using logarithm rules. $2\log_2 x < \log_2 (4x) + \log_2 (x + 3)$

Step 2: Use the properties of logarithms to combine the sum of the logarithms into a single logarithm. $2\log_2 x < \log_2 (4x \cdot (x + 3))$

Step 3: Remove the logarithm on the right side by taking both sides as the base of the logarithm. $2^{\left(2\log_2 x\right)} < 2^{\left(\log_2 (4x \cdot (x + 3))\right)}$

Step 4: Use the property of logarithms that $a^{\log_a (b)} = b$ to simplify both sides. $x^2 < 4x \cdot (x + 3)$

Step 5: Bring all the terms to one side of the inequality to solve for $x$. $x^2 - 4x \cdot (x + 3) < 0$

Step 6: Expand and simplify. $x^2 - 4x^2 - 12x < 0$

Step 7: Combine like terms. $-3x^2 - 12x < 0$

Step 8: Factor out the common term $-3x$. $-3x(x + 4) < 0$

Now, we have a quadratic inequality. To find the solution, we'll use a sign chart:

• Mark the critical points, where the expression on the left side becomes zero:

  • $x = 0$
  • $x + 4 = 0 \implies x = -4$

• Test the intervals determined by the critical points and see where the expression is positive or negative:

  • Test $x < -4$: Choose $x = -5$ (arbitrary value less than -4) $-3(-5)(-5 + 4) = 15 > 0$ (positive)
  • Test $-4 < x < 0$: Choose $x = -1$ (arbitrary value between -4 and 0) $-3(-1)(-1 + 4) = 9 > 0$ (positive)
  • Test $x > 0$: Choose $x = 1$ (arbitrary value greater than 0) $-3(1)(1 + 4) = -15 < 0$ (negative)

Step 9: Analyze the results from the sign chart:

• The inequality is satisfied when $x < -4$ or $-4 < x < 0$.
• The inequality is NOT satisfied when $x > 0$.

So, the solution to the inequality is $-4 < x < 0$.