If $\log_{10} 2 = a$ and $\log_{10} 3 = b$, express $\log_{5} 12$ in terms of $a$ and $b$.

Using the change of base formula, express $\log_5 12$ with base 10:

$$\log_5 12 = \frac{\log_{10} 12}{\log_{10} 5}$$

Simplify the numerator:

$$\log_{10} 12 = \log_{10}(2^2 \times 3) = 2\log_{10} 2 + \log_{10} 3 = 2a + b$$

Simplify the denominator:

$$\log_{10} 5 = \log_{10}\left(\frac{10}{2}\right) = \log_{10} 10 - \log_{10} 2 = 1 - a$$

Combine them back:

$$\log_5 12 = \frac{2a + b}{1 - a}$$

Answer: $\frac{2a + b}{1 - a}$