The problem asks to simplify the following logarithmic expression:

$\log_{10} 63 - 2 \log_{10} 7 + \log_{10} 6 - \log_{10} 72.$

Step 1: Apply logarithmic properties

Use these key logarithmic rules:

1. Product Rule: $\log_b (x) + \log_b (y) = \log_b (xy)$,
2. Quotient Rule: $\log_b (x) - \log_b (y) = \log_b \left(\frac{x}{y}\right)$,
3. Power Rule: $a \log_b (x) = \log_b (x^a)$.

Simplify step by step:

1. Rewrite $2 \log_{10} 7$ using the Power Rule: $2 \log_{10} 7 = \log_{10} (7^2) = \log_{10} 49.$

   The expression now becomes: $\log_{10} 63 - \log_{10} 49 + \log_{10} 6 - \log_{10} 72.$

2. Combine terms using the Quotient Rule: $\log_{10} 63 - \log_{10} 49 = \log_{10} \left(\frac{63}{49}\right),$ $\log_{10} 6 - \log_{10} 72 = \log_{10} \left(\frac{6}{72}\right).$

   The expression now becomes: $\log_{10} \left(\frac{63}{49}\right) + \log_{10} \left(\frac{6}{72}\right).$

3. Combine using the Product Rule: $\log_{10} \left(\frac{63}{49} \cdot \frac{6}{72}\right).$

Step 2: Simplify the argument

1. Simplify $\frac{63}{49}$: $\frac{63}{49} = \frac{9 \cdot 7}{7 \cdot 7} = \frac{9}{7}.$

2. Simplify $\frac{6}{72}$: $\frac{6}{72} = \frac{1}{12}.$

3. Combine the fractions: $\frac{63}{49} \cdot \frac{6}{72} = \frac{\frac{9}{7} \cdot \frac{1}{12}} = \frac{9}{84} = \frac{3}{28}.$

Step 3: Final expression

The expression simplifies to: $\log_{10} \left(\frac{3}{28}\right).$

This is the simplified form.

Would you like a more detailed breakdown or any clarification?

Related Questions:

1. What are other logarithmic properties and their applications?
2. How do you handle logarithms with different bases?
3. Can the value of $\log_{10} \frac{3}{28}$ be approximated?
4. What is the geometric interpretation of logarithms?
5. How can logarithms simplify real-world problems?

Tip:

Always check if logarithmic arguments can be further simplified to fractions or integers for easier computation.