To solve for $a \cdot b \cdot c \cdot d$ given the equations $4^a = 5$, $5^b = 6$, $6^c = 7$, and $7^d = 8$, we can take the natural logarithm of both sides of each equation.

Starting with $4^a = 5$:

$\ln(4^a) = \ln(5)$ $a \ln(4) = \ln(5)$ $a = \frac{\ln(5)}{\ln(4)}$

Next, for $5^b = 6$:

$\ln(5^b) = \ln(6)$ $b \ln(5) = \ln(6)$ $b = \frac{\ln(6)}{\ln(5)}$

Then, for $6^c = 7$:

$\ln(6^c) = \ln(7)$ $c \ln(6) = \ln(7)$ $c = \frac{\ln(7)}{\ln(6)}$

Finally, for $7^d = 8$:

$\ln(7^d) = \ln(8)$ $d \ln(7) = \ln(8)$ $d = \frac{\ln(8)}{\ln(7)}$

Now, we need to find $a \cdot b \cdot c \cdot d$:

$a \cdot b \cdot c \cdot d = \left( \frac{\ln(5)}{\ln(4)} \right) \left( \frac{\ln(6)}{\ln(5)} \right) \left( \frac{\ln(7)}{\ln(6)} \right) \left( \frac{\ln(8)}{\ln(7)} \right)$

Notice that in the product, all the logarithmic terms cancel each other out:

$a \cdot b \cdot c \cdot d = \frac{\ln(5)}{\ln(4)} \cdot \frac{\ln(6)}{\ln(5)} \cdot \frac{\ln(7)}{\ln(6)} \cdot \frac{\ln(8)}{\ln(7)} = \frac{\ln(8)}{\ln(4)}$

Simplifying the remaining expression:

$\frac{\ln(8)}{\ln(4)} = \frac{\ln(2^3)}{\ln(2^2)} = \frac{3 \ln(2)}{2 \ln(2)} = \frac{3}{2}$

Thus, the value of $a \cdot b \cdot c \cdot d$ is $\frac{3}{2}$.

Would you like more details or have any questions?

Here are five questions you might find interesting:

1. How do logarithmic properties simplify complex equations?
2. Can you explain the concept of natural logarithms?
3. How do you solve exponential equations with different bases?
4. What are some real-world applications of logarithms?
5. How can logarithmic identities help in solving integrals?

Tip: When dealing with exponential equations, taking the natural logarithm of both sides can greatly simplify the problem by reducing the exponent.