Logarithmic growth of an exponential Brownian clock with positive drift ($\mu >0⟹{\lim }_{t\to \infty }\frac{1}{t}\log {\int }_0^t{e}^{2B_s+2\mu s}ds=2\mu$)

The strong law for Brownian motion gives $B_s/s\to 0$. For every $0<\epsilon <\mu$, eventually $e^{(2\mu -2\epsilon )s}\le e^{2B_s+2\mu s}\le e^{(2\mu +2\epsilon )s}$. Integrating these bounds, taking logarithms and dividing by $t$ squeezes the growth rate between $2\mu -2\epsilon$ and $2\mu +2\epsilon$. Let $\epsilon \downarrow 0$. The corresponding inverse clock grows like $(\log t)/(2\mu )$.