Is $g(x)$, the first $(n -\log(n))$ bit of $f(x)$, a (strong) one-way function? Given a (strong) n-bit-by-n-bit one-way function $f$, is $g(x)$, the first $(n - \log(n))$ bit of $f(x)$, a (strong) one-way function, too?

When reading Prof. Sanjam Garg's Graduate Cryptography lecture notes, I came across this problem. I am trying to construct an adv. cracking $f$ based on an adv. cracking $g$, but unable to bound the ratio of $|\{x|g(x) = v$ and $f(x) = v||u\}|$ to $|\{x|g(x) = v\}|$ for arbitrary eligible $v$ and $u$ from the fact that $f$ is a one-way function.

Is there another way to prove/disprove this statement? Thanks.
Without knowing the details, note that any algorithm can utilize the $n-\log n$ bits of $g$ to invert it, can then brute force the remaining $\log n$ bits in time complexity $O(2^{\log n})=O(n),$ thus inverting $f.$ 

This will at most multiply the overall complexity by a linear factor, thus cannot achieve more than a polynomial gain between inverting $f$ and inverting $g$.