Functions between polynomial and exponential order

Since we can multiply $f$ by a positive constant and change nothing, assume $f(1)=1$ and let $g(x)= \log^+ f(x)$, strictly increasing from $1$ on and zero before so the hypothesis translates to $g(x)-k\log x$ unbounded for all positive $k$. We will show that for any $0<a<1$ it follows that $g(x)-g(ax)$ must be unbounded too so the answer to the question here is negative.

Assume by contradiction there is an $0<a<1, g(x)-g(ax) <M$ and let $n > -\frac{\log x}{\log a}$, so $a^nx<1, g(a^nx)=0$. Iterating the inequality above we get $g(x) < (n-1)M$ so since obviously we can replace integer $n$ with any higher real, so for example with $2c \log x$ where $c= -\frac{1}{\log a}$ constant , we get $g(x) < 2cM \log x$ and that contradicts the unboundness above for $k > 2cM$