Question: What is the value of $\log\left( \frac{x^2 y^3}{z} \right)$ when given the following:
$\log(x) = 3$
$\log(y) = 2$
$\log(z) = -1$
1. **State the problem:**
Find the value of $$\log\left( \frac{x^2 y^3}{z} \right)$$ given $$\log(x) = 3$$, $$\log(y) = 2$$, and $$\log(z) = -1$$.
2. **Recall the properties of logarithms:**
- $$\log\left( \frac{a}{b} \right) = \log(a) - \log(b)$$
- $$\log(a^n) = n \log(a)$$
- $$\log(ab) = \log(a) + \log(b)$$
3. **Apply the properties to the expression:**
$$\log\left( \frac{x^2 y^3}{z} \right) = \log(x^2 y^3) - \log(z)$$
Using the product rule:
$$\log(x^2 y^3) = \log(x^2) + \log(y^3)$$
Using the power rule:
$$\log(x^2) = 2 \log(x)$$
$$\log(y^3) = 3 \log(y)$$
4. **Substitute the given values:**
$$= 2 \log(x) + 3 \log(y) - \log(z)$$
$$= 2(3) + 3(2) - (-1)$$
5. **Calculate step-by-step:**
$$= 6 + 6 + 1$$
$$= 13$$
**Final answer:** $$13$$, which corresponds to option d.)