In general, the arbitrage-free price process $V_t$ at time $0 \le t \le T$ for a European claim $X =f(S_T)$ under the B-S model (which it looks like you have) is given by
$$V_t(X) = B_t\mathbb{E}_\mathbb{Q}[B_T^{-1}X | \mathcal{F}_t],$$
where $B_t$ is the bond price process, $\mathbb{Q}$ is the measure making the discounted B-S stock price process a $\mathbb{Q}$-martingale, and $\mathcal{F}_t$ is the sigma field for the Brownian Motion. This might be the "process" the question asks for, but you would need to evaluate this expectation at each $t$ to actually use it.
Here's an example of computing the initial value using an arbitrary payoff at maturity, $X = f(S_T)$, so all you'll have to do is replace the $f$ with your logarithm payoff.
I assume you're familiar with the B-S stock price process under the risk-neutral Brownian motion, $S_t = S_0\exp(\sigma \widetilde{W}_t + (r - \frac{1}{2}\sigma^2)t)$, where $\widetilde{W}_t$ is a $\mathcal{N}(0, t)$-distributed random variable for fixed $t$ under the $\mathbb{Q}$ measure, $r$ is the risk-free rate and $\sigma$, the volatility. Note the argument of the exponential function is a $\mathcal{N}((r - \frac{1}{2}\sigma^2)t, \sigma^2 t)$-distributed random variable for fixed $t$.
Define the random variable $Y$ ~ $\mathcal{N}(-\frac{1}{2}\sigma^2 T, \sigma^2 T)$. Then $S_T = S_0\exp(Y + rT)$, and letting $p(Y)$ denote the pdf of $Y$, from the formula for $V_t$ we have
$$V_0 = \mathbb{E}_\mathbb{Q}[\mathrm{e}^{-rT}f(S_0\exp(Y + rT))] \\ = \mathrm{e}^{-rT} \int_{-\infty}^\infty f(S_0\exp(y + rT))p(y)\mathrm{d}y \\ = \mathrm{e}^{-rT} \int_{-\infty}^\infty f(S_0\exp(y + rT))\frac{1}{\sqrt{2 \pi \sigma^2 T}}\exp\left(\frac{-(y + \frac{1}{2}\sigma^2T)^2}{2 \sigma^2 T}\right)\mathrm{d}y.$$