Suppose that $4^a = 5$, $5^b = 6$, $6^c = 7$, and $7^d = 8$. What is $a \cdot b \cdot c \cdot d$?
If I take the logarithm of both sides of each equation, using any base I want, I can simplify the exponents. For example, using base 4, I get $\log_4 (4^a) = \log_4 (5)$, which simplifies to $a = \log_4 (5)$. Similarly, using base 5, I get $b = \log_5 (6)$, using base 6, I get $c = \log_6 (7)$, and using base 7, I get $d = \log_7 (8)$. Now, to find the product $a \cdot b \cdot c \cdot d$, I can use the property of logarithms that says $\log_a (b) \cdot \log_b (c) = \log_a (c)$. For example, I can write $a \cdot b = \log_4 (5) \cdot \log_5 (6) = \log_4 (6)$. Similarly, I can write $a \cdot b \cdot c = \log_4 (6) \cdot \log_6 (7) = \log_4 (7)$. And finally, I can write $a \cdot b \cdot c \cdot d = \log_4 (7) \cdot \log_7 (8) = \log_4 (8)$. Now, I have simplified the product to a single logarithm, which I can evaluate. I know that $\log_4 (8) = \frac{3}{2}$, because $4^{\frac{3}{2}} = 2^3 = 8$. Therefore, the final answer is $\frac{3}{2}$. 