If f is outer, 1/f is outer (more generally f raised to any real - say non-zero to avoid constants - power is outer) - any outer function has no zeros as those are factored out with Blaschke products, so we can talk about log(f) and any complex power of f in the disk.

1/f always belongs to the Nevanlinna class N btw and if for example Re(f)>0, than Re(1/f)>0 so 1/f belongs to all hardy classes of exponent less than 1

Applying the above factorization with bounded analytic functions for class N for the ratio of any two outer functions, we get your answer

If $f$ is outer, $1/f$ is outer. (More generally $f$ raised to any real - say non-zero to avoid constants - power is outer.) Any outer function has no zeros as those are factored out with Blaschke products, so we can talk about $\log f$ and any complex power of $f$ in the disk.

$1/f$ always belongs to the Nevanlinna class $N$ by the way, and if for example $\Re f>0$, than $\Re(1/f)>0$ so $1/f$ belongs to all Hardy classes of exponent less than $1$.

Applying the above factorization with bounded analytic functions for class $N$ for the ratio of any two outer functions, we get your answer.