Question: 4) log \frac{(x+y)^3}{(x-y)^2}
log_a (x+y)^3 - log_a (x-y)^2
3 log_a (x+y) - 2 log_a (x-y)
5) \frac{\ln e^{3x} - (x+1)}{2x^2}
1. **Problem 4:** Simplify the logarithmic expression $$\log \frac{(x+y)^3}{(x-y)^2}$$.
2. **Formula used:** The logarithm of a quotient is the difference of logarithms: $$\log \frac{A}{B} = \log A - \log B$$.
3. Apply the quotient rule:
$$\log \frac{(x+y)^3}{(x-y)^2} = \log (x+y)^3 - \log (x-y)^2$$
4. Use the power rule of logarithms: $$\log A^n = n \log A$$.
5. Simplify:
$$3 \log (x+y) - 2 \log (x-y)$$
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6. **Problem 5:** Simplify the expression $$\frac{\ln e^{3x} - (x+1)}{2x^2}$$.
7. Recall that $$\ln e^{k} = k$$ for any $$k$$.
8. So, $$\ln e^{3x} = 3x$$.
9. Substitute back:
$$\frac{3x - (x+1)}{2x^2} = \frac{3x - x - 1}{2x^2} = \frac{2x - 1}{2x^2}$$
10. Express the fraction:
$$\frac{2x - 1}{2x^2} = \frac{\cancel{2}x - 1}{\cancel{2}x^2}$$ (no common factor to cancel fully, so leave as is).
**Final answers:**
- For problem 4: $$3 \log (x+y) - 2 \log (x-y)$$
- For problem 5: $$\frac{2x - 1}{2x^2}$$