Statement
When two logarithms with the same base ($ a $) are added, the inside parts are multiplied:
$$ \log_a(b) + \log_a(c) = \log_a(bc) $$
Proof
Given the sum $ \log_a(b) + \log_a(c) $, let $ x $ and $ y $ be the terms of the sum. By the definition of the logarithm, the following applies:
$$ x = \log_a(b) \implies a^x = b $$
$$ y = \log_a(c) \implies a^y = c $$
Now multiply $ b $ and $ c $ and add the exponents.
$$ b * c = a^x * a^y = a^{x + y} $$
Take the logarithm with base $ a $ on both sides.
$$ \log_a(b * c) = \log_a(a^{x + y}) $$
Simplify the second logarithm, since $ \log_a(...) $ and $ a^{...} $ are inverse operations.
$$ \log_a(bc) = x + y $$
Finally, re-substitute $ x $ and $ y $ and flip the equation.
$$ \log_a(bc) = \log_a(b) + \log_a(c) $$
$$ \log_a(b) + \log_a(c) = \log_a(bc) $$