Problem 2. Consider the set $G=(2 ; \infty)$ and the function $f: G \rightarrow \mathbb{R}, f(x)=\ln (x-2)$. Determine a composition law * on $G$, such that $\left(G,{ }^{*}\right)$ is a group, and the function $f$ is an isomorphism from the group $\left(G,{ }^{*}\right)$ to the additive group $(\mathbb{R},+)$.
Solution:

Since the function $\mathrm{f}$ is an isomorphism from the group $\left(G,{ }^{*}\right)$ to the additive group $(\mathbb{R},+)$, we obtain that $f$ is bijective, hence invertible (1 p)

and the relation $f(x * y)=f(x)+f(y)$, for all $x, y \in G$. (1 p)

$$
\begin{aligned}
& f^{-1}(f(x * y))=f^{-1}(f(x)+f(y)), \text { for all } x, y \in G \\
& x * y=f^{-1}(f(x)+f(y))
\end{aligned}
$$

It is shown that $f^{-1}: \mathbb{R}_{+}^{*} \rightarrow G, f^{-1}(x)=e^{x}+2(\mathbf{1} \mathbf{p})$

and the relation above becomes

$$
\begin{aligned}
& x * y=e^{f(x)+f(y)}+2=e^{\ln (x-2)+\ln (y-2)}+2=e^{\ln (x-2)(y-2)}+2 \quad(1 \mathbf{p}) \\
& =(x-2)(y-2)+2 \\
& x * y=x y-2 x-2 y+6, \text { for all } x, y \in(2 ; \infty) \quad \text { (1 p) }
\end{aligned}
$$

It is shown that * is a binary operation on $G$ and that $\left(G,{ }^{*}\right)$ is a group.