To solve the expression

$\frac{\log_{a}b}{\log_{ac}b}-\frac{\log_{a}d}{\log_{c}d}$

let's first break down each term.

1. Simplify $\frac{\log_{a}b}{\log_{ac}b}$

We can use the change of base formula, $\log_{a}b = \frac{\log b}{\log a}$, to rewrite each term:

$\log_{a}b = \frac{\log b}{\log a}$ $\log_{ac} b = \frac{\log b}{\log(ac)} = \frac{\log b}{\log a + \log c}$

So,

$\frac{\log_{a}b}{\log_{ac}b} = \frac{\frac{\log b}{\log a}}{\frac{\log b}{\log a + \log c}} = \frac{\log b}{\log a} \cdot \frac{\log a + \log c}{\log b}$

This simplifies to:

$\frac{\log a + \log c}{\log a} = 1 + \frac{\log c}{\log a} = 1 + \log_{a}c$

2. Simplify $\frac{\log_{a}d}{\log_{c}d}$

Similarly, using the change of base formula:

$\log_{a}d = \frac{\log d}{\log a}$ $\log_{c}d = \frac{\log d}{\log c}$

So,

$\frac{\log_{a}d}{\log_{c}d} = \frac{\frac{\log d}{\log a}}{\frac{\log d}{\log c}} = \frac{\log d}{\log a} \cdot \frac{\log c}{\log d} = \frac{\log c}{\log a} = \log_{a}c$

3. Substitute into the original expression

Now substitute the simplified terms back into the original expression:

$\left(1 + \log_{a}c\right) - \log_{a}c$

This simplifies to:

$1$

Hence, the value of the entire expression is 1, regardless of the values of $a$, $b$, $c$, and $d$ as long as the conditions $a > 1$, $b > 1$, $c > 1$, and $d > 1$ are met.

Questions for further understanding:

1. How does the change of base formula work in logarithms?
2. What are the conditions under which logarithmic properties can be applied?
3. How would the expression change if we had $a, b, c,$ or $d \leq 1$?
4. What would be the result if one of the variables $a, b, c,$ or $d$ was unknown?
5. Can this result be generalized to expressions with different base logarithms?

Tip: When dealing with complex expressions involving logarithms, using the change of base formula can greatly simplify the problem.