Logarithm on Positive Real Numbers is Group Isomorphism

Theorem
Let $\left({\R_{>0}, \times}\right)$ be the multiplicative group of positive real numbers.

Let $\left({\R, +}\right)$ be the additive group of real numbers.

Let $b$ be any real number such that $b > 1$.

Let $\log_b: \left({\R_{>0}, \times}\right) \to \left({\R, +}\right)$ be the mapping:
 * $x \mapsto \log_b \left({x}\right)$

where $\log_b$ is the logarithm to base $b$.

Then $\log_b$ is a group isomorphism.

Proof
From Sum of Logarithms we have:
 * $\forall x, y \in \R_{>0}: \log_b \left({x y}\right) = \log_b x + \log_b y$

That is $\log_b$ is a group homomorphism.

From Change of Base of Logarithm, $\log_b$ is a constant multiplied by the natural logarithm function.

Then we have that Logarithm is Strictly Increasing and Strictly Concave.

From Strictly Monotone Function is Bijective, it follows that $\log_b$ is a bijection.

So $\log_b$ is a bijective group homomorphism, and so a group isomorphism.