# The sum of logarithms

## 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.

$$\log_a(bc) = \log_a(b) + \log_a(c)$$

$$\log_a(b) + \log_a(c) = \log_a(bc)$$

## Proofs building upon this proof

### The difference of logarithms

This proof shows that the difference of two logarithms with the same base is just one logarithm with the inside parts being divided.