Question: Given that $5^p = 9$, $9^q = 12$, $12^r = 16$, $16^s = 20$ and $20^t = 25$, find the value of $pqrst$.
1. **State the problem:** We are given the equations $5^p = 9$, $9^q = 12$, $12^r = 16$, $16^s = 20$, and $20^t = 25$. We need to find the value of the product $pqrst$.
2. **Express each variable in terms of logarithms:**
From $5^p = 9$, taking logarithm base 5, we get:
$$p = \log_5 9$$
Similarly,
$$q = \log_9 12$$
$$r = \log_{12} 16$$
$$s = \log_{16} 20$$
$$t = \log_{20} 25$$
3. **Recall the change of base formula:**
$$\log_a b = \frac{\log b}{\log a}$$
where the logarithm base on the right side can be any convenient base (e.g., natural log or base 10).
4. **Write the product $pqrst$ using change of base:**
$$pqrst = \left(\frac{\log 9}{\log 5}\right) \left(\frac{\log 12}{\log 9}\right) \left(\frac{\log 16}{\log 12}\right) \left(\frac{\log 20}{\log 16}\right) \left(\frac{\log 25}{\log 20}\right)$$
5. **Simplify the expression by canceling common terms:**
$$pqrst = \frac{\cancel{\log 9}}{\log 5} \times \frac{\log 12}{\cancel{\log 9}} \times \frac{\log 16}{\log 12} \times \frac{\log 20}{\log 16} \times \frac{\log 25}{\log 20}$$
Canceling all intermediate terms, we get:
$$pqrst = \frac{\log 25}{\log 5}$$
6. **Evaluate the remaining logarithm:**
Since $25 = 5^2$, we have:
$$\log 25 = \log (5^2) = 2 \log 5$$
Therefore,
$$pqrst = \frac{2 \log 5}{\log 5} = 2$$
**Final answer:**
$$\boxed{2}$$