Consider a random sample of size n n that follows a density probability function given...

Indeed you can use logarithms if you want to, and you don’t need to consider $\log f$ instead of $f$ in order to do so. Since
$f(x\mid \theta )=\prod _{i=1}^{n}f(x_i,\theta )=\frac{1}{{\theta }^n}\prod _{i=1}^n{x}_i^{-\frac{\theta +1}{\theta }}=\frac{1}{{\theta }^n}\exp (-\frac{\theta +1}{\theta }\sum _{i=1}^n\ln x_i),$
we have
$f(x\mid \theta )=g(T(x)\mid \theta )h(x),$
with
$T(x)=\sum _{i=1}^n\ln x.$
Thus $T(x)$ is a sufficient statistic.
But you don’t have to use logarithms. You could also just write
$f(x\mid \theta )=\prod _{i=1}^{n}f(x_i,\theta )=\frac{1}{{\theta }^n}\prod _{i=1}^n{x}_i^{-\frac{\theta +1}{\theta }}=\frac{1}{{\theta }^n}{\left(\prod _{i=1}^n{x}_i\right)}^{-\frac{\theta +1}{\theta }},$
which factorizes with $S(x)=\prod _{i=1}^n{x}_i$. That $S(x)$ and $T(x)$ are likewise sufficient statistics is not surprising, since one is an injective function of the other$S(x)=\exp (T(x))$ and $T(x)=\log (S(x))$, and an injective function of a sufficient statistic is again a sufficient statistic.