Consider to prove the result, first prove f(x) = O(g(x)) and then prove g(x) = O(f(x)).

Steps to Solve

1. Proving $f(x) = O(g(x))$

We need to show that there exist constants $C$ and $k$ such that for all $x > k$, the inequality $|f(x)| \leq C |g(x)|$ holds.

Starting with $f(x) = \log(x^2 + 1)$ and $g(x) = \log(x)$:

Observe that for $x > k$, we can bound $f(x)$ as follows:

$$ x^2 + 1 \leq 2x^2 \quad \text{for} , x > 1 $$

Taking the logarithm, we get:

$$ \log(x^2 + 1) \leq \log(2x^2) = \log(2) + 2\log(x) $$

Thus, we can set $C = 2 + \log(2)$ and $k = 1$.

2. Establishing the first inequality

From the previous step, we have:

$$ f(x) \leq C g(x) \quad \text{for} ; x > k $$

So we can conclude that $f(x) = O(g(x))$.

3. Proving $g(x) = O(f(x))$

Now we need to show that there exist constants $C'$ and $k'$ such that for all $x > k'$, the inequality $|g(x)| \leq C' |f(x)|$ holds.

Since $g(x) = \log(x)$:

For $x > k'$, we can find a bound for $g(x)$:

$$ \log(x) \leq \log(x^2 + 1) \quad \text{for} ; x > 1 $$

This can be shown as:

$$ x^2 + 1 \geq x^2 \implies \log(x^2 + 1) \geq \log(x^2) $$

So we have:

$$ g(x) \leq f(x) \quad \text{for} ; x > k' $$

4. Finishing up the proof

Combining the results from both parts, since $f(x) = O(g(x))$ and $g(x) = O(f(x))$, we conclude that:

$$ f(x) \sim g(x) \quad \text{as} ; x \to \infty $$

Thus, we have established that $f(x)$ and $g(x)$ grow at the same rate asymptotically.