Answer :
Sure, let's break down the given inequality and solve it step-by-step. The inequality we have is:
[tex]\[ \log_2(x + 3) - \log_2(x) + \log_2(2x^2) \leq 3 \][/tex]
First, use the properties of logarithms to simplify the expression.
Step 1: Combine the logarithmic terms.
We apply the properties of logarithms:
- [tex]\(\log_b(a) - \log_b(b) = \log_b\left(\frac{a}{b}\right)\)[/tex]
- [tex]\(\log_b(a) + \log_b(b) = \log_b(ab)\)[/tex]
Applying these properties:
[tex]\[ \log_2(x+3) - \log_2(x) = \log_2\left(\frac{x+3}{x}\right) \][/tex]
[tex]\[ \log_2(2x^2) = \log_2(2) + \log_2(x^2) = 1 + 2\log_2(x) \][/tex]
Combining all the logarithmic terms gives us:
[tex]\[ \log_2\left(\frac{x+3}{x}\right) + 1 + 2\log_2(x) \leq 3 \][/tex]
Step 2: Further simplify the inequality.
Substitute back into the inequality:
[tex]\[ \log_2\left(\frac{x+3}{x}\right) + 1 + 2\log_2(x) \leq 3 \][/tex]
Combine the constant terms:
[tex]\[ \log_2\left(\frac{x+3}{x}\right) + 2\log_2(x) \leq 2 \][/tex]
Step 3: Combine the logarithms under a single logarithm term.
Use the property:
[tex]\(\log_b(a) + \log_b(b) = \log_b(ab)\)[/tex]
[tex]\[ \log_2\left(\frac{x+3}{x} \cdot x^2\right) \leq 2 \][/tex]
Simplify the inside of the logarithm:
[tex]\[ \log_2\left(\frac{x+3}{x} \cdot x^2\right) = \log_2(x^2 + 3x) \][/tex]
Step 4: Exponentiate both sides to remove the logarithm term.
We have:
[tex]\[ \log_2(x^2 + 3x) \leq 2 \][/tex]
By exponentiating both sides with base 2, we get:
[tex]\[ x^2 + 3x \leq 2^2 \][/tex]
Simplify the exponentiation:
[tex]\[ x^2 + 3x \leq 4 \][/tex]
Step 5: Solve the quadratic inequality.
Rewrite the inequality in standard quadratic form:
[tex]\[ x^2 + 3x - 4 \leq 0 \][/tex]
Factor the quadratic equation:
[tex]\[ (x + 4)(x - 1) \leq 0 \][/tex]
Step 6: Solve the factored inequality.
The critical points are [tex]\(x = -4\)[/tex] and [tex]\(x = 1\)[/tex]. These points divide the number line into three intervals: [tex]\( (-\infty, -4) \)[/tex], [tex]\( (-4, 1) \)[/tex], and [tex]\( (1, \infty) \)[/tex].
Determine where the quadratic expression is non-positive by testing values from each interval:
1. [tex]\( x < -4\)[/tex] (e.g., [tex]\( x = -5 \)[/tex]):
[tex]\[ (-5 + 4)(-5 - 1) = (-1)(-6) = 6 \quad (\text{positive}) \][/tex]
2. [tex]\(-4 \leq x \leq 1 \)[/tex]:
[tex]\[ (x + 4)(x - 1) \leq 0 \quad (\text{interval satisfies the inequality}) \][/tex]
3. [tex]\( x > 1 \)[/tex] (e.g., [tex]\( x = 2 \)[/tex]):
[tex]\[ (2 + 4)(2 - 1) = 6 \quad (\text{positive}) \][/tex]
Thus, the inequality [tex]\( (x + 4)(x - 1) \leq 0 \)[/tex] holds for:
[tex]\[-4 \leq x \leq 1\][/tex]
Step 7: Verify the domain constraints.
The original logarithmic expressions impose constraints:
[tex]\[ x + 3 > 0 \quad \text{and} \quad x > 0 \implies x > 0 \][/tex]
Thus, [tex]\(x > 0\)[/tex] adds an additional constraint, limiting the solution to:
[tex]\[0 < x \leq 1\][/tex]
Final Solution:
The solution to the inequality [tex]\(\log_2(x + 3) - \log_2(x) + \log_2(2x^2) \leq 3\)[/tex] is:
[tex]\[ x \in (0, 1] \][/tex]
[tex]\[ \log_2(x + 3) - \log_2(x) + \log_2(2x^2) \leq 3 \][/tex]
First, use the properties of logarithms to simplify the expression.
Step 1: Combine the logarithmic terms.
We apply the properties of logarithms:
- [tex]\(\log_b(a) - \log_b(b) = \log_b\left(\frac{a}{b}\right)\)[/tex]
- [tex]\(\log_b(a) + \log_b(b) = \log_b(ab)\)[/tex]
Applying these properties:
[tex]\[ \log_2(x+3) - \log_2(x) = \log_2\left(\frac{x+3}{x}\right) \][/tex]
[tex]\[ \log_2(2x^2) = \log_2(2) + \log_2(x^2) = 1 + 2\log_2(x) \][/tex]
Combining all the logarithmic terms gives us:
[tex]\[ \log_2\left(\frac{x+3}{x}\right) + 1 + 2\log_2(x) \leq 3 \][/tex]
Step 2: Further simplify the inequality.
Substitute back into the inequality:
[tex]\[ \log_2\left(\frac{x+3}{x}\right) + 1 + 2\log_2(x) \leq 3 \][/tex]
Combine the constant terms:
[tex]\[ \log_2\left(\frac{x+3}{x}\right) + 2\log_2(x) \leq 2 \][/tex]
Step 3: Combine the logarithms under a single logarithm term.
Use the property:
[tex]\(\log_b(a) + \log_b(b) = \log_b(ab)\)[/tex]
[tex]\[ \log_2\left(\frac{x+3}{x} \cdot x^2\right) \leq 2 \][/tex]
Simplify the inside of the logarithm:
[tex]\[ \log_2\left(\frac{x+3}{x} \cdot x^2\right) = \log_2(x^2 + 3x) \][/tex]
Step 4: Exponentiate both sides to remove the logarithm term.
We have:
[tex]\[ \log_2(x^2 + 3x) \leq 2 \][/tex]
By exponentiating both sides with base 2, we get:
[tex]\[ x^2 + 3x \leq 2^2 \][/tex]
Simplify the exponentiation:
[tex]\[ x^2 + 3x \leq 4 \][/tex]
Step 5: Solve the quadratic inequality.
Rewrite the inequality in standard quadratic form:
[tex]\[ x^2 + 3x - 4 \leq 0 \][/tex]
Factor the quadratic equation:
[tex]\[ (x + 4)(x - 1) \leq 0 \][/tex]
Step 6: Solve the factored inequality.
The critical points are [tex]\(x = -4\)[/tex] and [tex]\(x = 1\)[/tex]. These points divide the number line into three intervals: [tex]\( (-\infty, -4) \)[/tex], [tex]\( (-4, 1) \)[/tex], and [tex]\( (1, \infty) \)[/tex].
Determine where the quadratic expression is non-positive by testing values from each interval:
1. [tex]\( x < -4\)[/tex] (e.g., [tex]\( x = -5 \)[/tex]):
[tex]\[ (-5 + 4)(-5 - 1) = (-1)(-6) = 6 \quad (\text{positive}) \][/tex]
2. [tex]\(-4 \leq x \leq 1 \)[/tex]:
[tex]\[ (x + 4)(x - 1) \leq 0 \quad (\text{interval satisfies the inequality}) \][/tex]
3. [tex]\( x > 1 \)[/tex] (e.g., [tex]\( x = 2 \)[/tex]):
[tex]\[ (2 + 4)(2 - 1) = 6 \quad (\text{positive}) \][/tex]
Thus, the inequality [tex]\( (x + 4)(x - 1) \leq 0 \)[/tex] holds for:
[tex]\[-4 \leq x \leq 1\][/tex]
Step 7: Verify the domain constraints.
The original logarithmic expressions impose constraints:
[tex]\[ x + 3 > 0 \quad \text{and} \quad x > 0 \implies x > 0 \][/tex]
Thus, [tex]\(x > 0\)[/tex] adds an additional constraint, limiting the solution to:
[tex]\[0 < x \leq 1\][/tex]
Final Solution:
The solution to the inequality [tex]\(\log_2(x + 3) - \log_2(x) + \log_2(2x^2) \leq 3\)[/tex] is:
[tex]\[ x \in (0, 1] \][/tex]
