Let's analyze and solve question 6:

The problem states:

$\frac{\log a}{b-c} = \frac{\log b}{c-a} = \frac{\log c}{a-b}.$

We need to find the product $a \cdot b \cdot c$.

Step 1: Represent the common ratio

Let the common value of the given ratios be $k$. Thus, we can write: $\frac{\log a}{b-c} = k, \quad \frac{\log b}{c-a} = k, \quad \frac{\log c}{a-b} = k.$

From these, we get: $\log a = k(b-c),$ $\log b = k(c-a),$ $\log c = k(a-b).$

Step 2: Add the equations

Adding all three equations: $\log a + \log b + \log c = k(b-c) + k(c-a) + k(a-b).$

On the right-hand side, the terms $k(b-c) + k(c-a) + k(a-b)$ cancel out because they sum to zero: $k(b-c + c-a + a-b) = k(0) = 0.$

Thus: $\log a + \log b + \log c = 0.$

Step 3: Simplify using properties of logarithms

The sum of logarithms is the logarithm of the product: $\log (a \cdot b \cdot c) = 0.$

Exponentiating both sides: $a \cdot b \cdot c = 10^0 = 1.$

Final Answer:

$a \cdot b \cdot c = 1.$

Let me know if you want further clarification or details!

Related Questions:

1. What are the properties of logarithms used here?
2. How do we ensure the given conditions of equality are consistent?
3. Can you solve this using substitution instead of a general $k$?
4. What happens if $\log a = \log b = \log c$?
5. What are other applications of symmetric ratios in equations like this?

Tip: When multiple ratios are given as equal, introduce a common parameter to simplify relationships.