We'll show two results. The first is that for any positive constant $k$, we'll have
$$
k\cdot f(n) = O(f(n))
$$
To see this, recall what it means for $g(n)=O(f(n))$: this is true when we can find two constants $c>0, N\ge 0$ such that $g(n)\le c\cdot f(n)$ for all $n\ge N$. If we choose $g(n)$ to be $k\cdot f(n)$ then we have just what we need: we'll just use $k$ for our constant $c$ and select any $N$ we want. It's trivial to adapt this argument to big-$\Omega$, and so it also holds for big-$\Theta$ as well.
In simple terms: Asymptotic notation ignores constant multiples.
The second result we need is that logs to different bases are always going to be constant multiples of each other, namely
$$
\log_ax=(\log_ab)\log_bx
$$
This is easy to see: just raise $a$ to each side power and you'll wind up with equal results.
In the problem stated, assuming that $\lg n$ is $\log_2n$, we'll have
$$
\lg n=(\log_2 8)\log_8n=3\log_8n
$$
and the two logs will differ only by a multiple of 3, which as we showed above, means that $\lg n$ and $\log_8n$ are the same function, as far as big-whatever notation is concerned.
