Question: given that $\log_a y = k$ express each of the following in terms of $k$ a) $(\log_a y)^3$ b) $\log_a y^3$ c) $\log_a (ay)^2$ d) $\log_a \frac{\sqrt{y}}{a^2}$
1. **State the problem:** Given $\log_a y = k$, express each expression in terms of $k$.
2. **Recall logarithm rules:**
- Power rule: $\log_a (x^m) = m \log_a x$
- Product rule: $\log_a (xy) = \log_a x + \log_a y$
- Quotient rule: $\log_a \left(\frac{x}{y}\right) = \log_a x - \log_a y$
- Root as power: $\sqrt{y} = y^{\frac{1}{2}}$
3. **Solve each part:**
**a) $(\log_a y)^3$**
Since $\log_a y = k$, then
$$ (\log_a y)^3 = k^3 $$
**b) $\log_a y^3$**
Using power rule:
$$ \log_a y^3 = 3 \log_a y = 3k $$
**c) $\log_a (ay)^2$**
Using product and power rules:
$$ \log_a (ay)^2 = 2 \log_a (ay) = 2 (\log_a a + \log_a y) $$
Since $\log_a a = 1$ and $\log_a y = k$:
$$ = 2 (1 + k) = 2 + 2k $$
**d) $\log_a \frac{\sqrt{y}}{a^2}$**
Rewrite numerator and denominator:
$$ \log_a \left(\frac{y^{\frac{1}{2}}}{a^2}\right) = \log_a y^{\frac{1}{2}} - \log_a a^2 $$
Apply power rule:
$$ = \frac{1}{2} \log_a y - 2 \log_a a = \frac{1}{2} k - 2 \cdot 1 = \frac{k}{2} - 2 $$
4. **Final answers:**
- a) $k^3$
- b) $3k$
- c) $2 + 2k$
- d) $\frac{k}{2} - 2$