1. **Solve** $3 \log_a - \log_a (x+2)$. - The problem involves simplifying logarithmic expressions. - Recall the logarithm subtraction rule: $\log_a M - \log_a N = \log_a \left(\frac{M}{N}\right)$. - Also, $k \log_a M = \log_a M^k$. Step 1: Rewrite $3 \log_a$ as $\log_a a^3$ because $\log_a a = 1$. Step 2: Apply the subtraction rule: $$3 \log_a - \log_a (x+2) = \log_a a^3 - \log_a (x+2) = \log_a \left(\frac{a^3}{x+2}\right)$$ --- 2. **Find 4 in terms of** $\log_a (x-2) = \log_a (x+2)$. - The equation $\log_a (x-2) = \log_a (x+2)$ implies the arguments are equal (since $\log_a$ is one-to-one): $$x-2 = x+2$$ - This is impossible unless $2 = -2$, which is false. - Therefore, no solution exists for $x$ in this equation. - The problem asks to "Find 4 in terms of" this equation, which is unclear; assuming it means to express 4 using these logs, but since the logs are equal only if $x-2 = x+2$, no valid $x$ exists. --- 3. **Solve** $5 \log_a - 2 \log_a (x+4) = 2 \log_a Y + \log_a$. - Rewrite terms: - $5 \log_a = \log_a a^5$ - $2 \log_a (x+4) = \log_a (x+4)^2$ - $2 \log_a Y = \log_a Y^2$ - $\log_a = \log_a a$ - Substitute: $$\log_a a^5 - \log_a (x+4)^2 = \log_a Y^2 + \log_a a$$ - Use log subtraction and addition rules: $$\log_a \left(\frac{a^5}{(x+4)^2}\right) = \log_a (a Y^2)$$ - Since logs are equal, their arguments are equal: $$\frac{a^5}{(x+4)^2} = a Y^2$$ - Divide both sides by $a$: $$\frac{a^4}{(x+4)^2} = Y^2$$ - Take square root: $$Y = \pm \frac{a^2}{x+4}$$ --- 4. **Divide** $(x^2 + 3x + 4)$ by $(2x + 5)$. - Polynomial division: Step 1: Divide leading terms: $x^2 \div 2x = \frac{x}{2}$. Step 2: Multiply divisor by $\frac{x}{2}$: $2x \times \frac{x}{2} = x^2$, $5 \times \frac{x}{2} = \frac{5x}{2}$. Step 3: Subtract: $(x^2 + 3x + 4) - (x^2 + \frac{5x}{2}) = 3x - \frac{5x}{2} + 4 = \left(3 - \frac{5}{2}\right)x + 4 = \frac{1}{2}x + 4$. Step 4: Divide leading terms: $\frac{1}{2}x \div 2x = \frac{1}{4}$. Step 5: Multiply divisor by $\frac{1}{4}$: $2x \times \frac{1}{4} = \frac{1}{2}x$, $5 \times \frac{1}{4} = \frac{5}{4}$. Step 6: Subtract: $(\frac{1}{2}x + 4) - (\frac{1}{2}x + \frac{5}{4}) = 4 - \frac{5}{4} = \frac{11}{4}$. - Quotient: $\frac{x}{2} + \frac{1}{4}$, remainder $\frac{11}{4}$. - Final answer: $$\frac{x^2 + 3x + 4}{2x + 5} = \frac{x}{2} + \frac{1}{4} + \frac{\frac{11}{4}}{2x + 5}$$ --- 5. **Divide** $(12x^3 - 2x^2 - 3x + 28)$ by $(3x + 4)$. - Polynomial division: Step 1: Divide leading terms: $12x^3 \div 3x = 4x^2$. Step 2: Multiply divisor by $4x^2$: $3x \times 4x^2 = 12x^3$, $4 \times 4x^2 = 16x^2$. Step 3: Subtract: $(12x^3 - 2x^2) - (12x^3 + 16x^2) = -18x^2$. Step 4: Bring down $-3x$: new dividend $-18x^2 - 3x$. Step 5: Divide leading terms: $-18x^2 \div 3x = -6x$. Step 6: Multiply divisor by $-6x$: $3x \times -6x = -18x^2$, $4 \times -6x = -24x$. Step 7: Subtract: $(-18x^2 - 3x) - (-18x^2 - 24x) = 21x$. Step 8: Bring down $+28$: new dividend $21x + 28$. Step 9: Divide leading terms: $21x \div 3x = 7$. Step 10: Multiply divisor by $7$: $3x \times 7 = 21x$, $4 \times 7 = 28$. Step 11: Subtract: $(21x + 28) - (21x + 28) = 0$. - Quotient: $4x^2 - 6x + 7$, remainder $0$. - Final answer: $$\frac{12x^3 - 2x^2 - 3x + 28}{3x + 4} = 4x^2 - 6x + 7$$