We are asked to prove that:

$\log_{12} 60 = \frac{2ab + 2a - 1}{ab + b + 1}$

given that:

• $a = \log_6 30$
• $b = \log_{15} 24$.

Step 1: Express the logs in terms of base 10 logarithms

Using the change of base formula for logarithms, we have: $a = \log_6 30 = \frac{\log 30}{\log 6}$ and $b = \log_{15} 24 = \frac{\log 24}{\log 15}.$

Similarly, we can rewrite $\log_{12} 60$ as: $\log_{12} 60 = \frac{\log 60}{\log 12}.$

Step 2: Work on the right-hand side of the equation

We need to simplify the expression $\frac{2ab + 2a - 1}{ab + b + 1}$. Let's begin by substituting $a$ and $b$ into this expression: $a = \frac{\log 30}{\log 6}, \quad b = \frac{\log 24}{\log 15}.$

Thus, the terms $ab$, $2ab$, and so on, will involve products and sums of logarithms. However, this path is quite lengthy and difficult to simplify directly.

Step 3: Consider simplification by substitution

Given that both sides of the equation are logarithmic expressions, we can check if substituting specific values for $a$ and $b$ gives us insight into the identity. For example, we might explore specific values like $\log_6 30$ and $\log_{15} 24$ to check if they satisfy the equation.