Proving that given metric space is complete: $X := (0,\infty)$ and $d:=|\ln(x)-\ln(y)|$

Given the metric space $(X,d)$ with $X := (0,\infty)$ and $d:=|\ln(x)-\ln(y)|$, how can I show that $(X,d)$ is complete?

I need to prove that any Cauchy sequence converges, so:

If $(x_n)$ is a Cauchy sequence in X, then it follows for all $\epsilon \gt 0$ that there exists an $n_0 \in \mathbb{R} : \forall n,m \leq n_0: d(x_n,x_m) \lt \epsilon$.

I couldn't find a direct way to prove this, I guess an indirect approach might go as follows:

Let $(x_n)$ be a Cauchy sequence in X and assume that it does not converge, then it follows that there exists an $\epsilon > 0$ such that for an arbitrary $x \in X : \forall n \in \mathbb{N}: d(x_n - x) \geq \epsilon$ and this would contradict the fact that $(x_n)$ is a cauchy sequence. I'm assuming that this is incorrect since I didn't even use the given metric and this proof would mean that no cauchy sequence converges.

So how can I prove that this metric space is complete? And in general is there a way how to approach these completeness proofs?

HINT: If $x_n\to x_0$ in the standard metric on $(0,\infty)$, then $\ln x_n \to \ln x_0$.
• Is this only in the direction $x_n \rightarrow x \Rightarrow \ln x_n \rightarrow x$ or is this an iff statement? – eager2learn Jul 23 '15 at 13:23
• Yes, $(0,\infty)$ and $\bf R$ are homeomorphic (do you know this definition?) by function $\ln$, because it is one-to-one and both $\ln$ and $\exp$ are continuous. – Przemysław Scherwentke Jul 23 '15 at 13:27
• Yes I know the definition. Ok so I could simply use the fact that any Cauchy sequence in $\mathbb{R}$ with the standard metric converges and the fact that ln is homeomorphic and that would allow me to prove that it also converges with respect to this given metric, right? – eager2learn Jul 23 '15 at 13:40
Let $(x_n)$ be a Cauchy sequence then $d(x_n, x_m)<\epsilon$ so $\mid \ln(x_n)-\ln(x_m)\mid <\epsilon.$ But $$\mid \ln(x_n)-\ln(x_m)\mid=\mid \ln\frac{x_n}{x_m}\mid<\epsilon$$ So $\frac{x_n}{x_m} \rightarrow 1$ then subsequencec $(x_n)$ and $(x_m)$ have same limit and the sequence is convergent.
Hint If $x_n$ is Cauchy, prove that $e^{x_n}$ is Cauchy with the usual metric of $\mathbb R$. If $y$ is the limit of $e^{x_n}$ then.....