A player can bet a quantity $u_k\geq0$, at each instant $k$ if $u_k \leq x_k$, where $x_k$ is his current fortune at instant $k$.
He wins the money he bets with probability $\frac{1}{2}<p <1$ or loses this money with probability $1-p$.
What's the strategy that this player can use to maximize $E(\ln(x_N))$, where $x_N$ represents his fortune at instant $N$?
We will assume there is no limit to the precision with which the player can divide their money, then the problem is self-similar and the answer will be the same each iteration. Let $\left(a_i\right)$ be the tuple of successive proportions betted. In each iteration $i$,$$ \begin{align} x_i=x_{i-1}*\left\{\begin{array}{cl} \left(1+a_i\right) & : p \\ \left(1-a_i\right) & : 1-p \end{array} \right.\\ \iff\log\left(x_i\right)=\log\left(x_{i-1}\right)+\left\{\begin{array}{cl} \log\left(1+a_i\right) & : p \\ \log\left(1-a_i\right) & : 1-p \end{array} \right. \\ E\left(\log\left(x_i\right)\right)=E\left(\log\left(x_{i-1}\right)\right)+p*\log\left(1+a_i\right)+(1-p)*\log\left(1-a_i\right) \end{align}$$So each iteration we will choose $a_i$ to maximise $p*\log\left(1+a_i\right)+(1-p)*\log\left(1-a_i\right)$. Note that we may take its $\exp$ becaus it's over the positive numbers,$$ \begin{align} \exp\left(p*\log\left(1+a_i\right)+(1-p)*\log\left(1-a_i\right)\right)\\ =\left(1+a_i\right)^p*\left(1-a_i\right)^{1-p} \end{align} $$At its maximum, its derivative will be $0$,$$ \begin{align} \mathrm{diff}_{a_i}\left(\left(1+a_i\right)^p*\left(1-a_i\right)^{1-p}\right)\\ =\left(1-a_i\right)^{1-p}*\mathrm{diff}_{a_i}\left(\left(1+a_i\right)^p\right)+\left(1+a_i\right)^p*\mathrm{diff}_{a_i}\left(\left(1-a_i\right)^{1-p}\right)\\ =\left(1-a_i\right)^{1-p}*p*\left(1+a_i\right)^{p-1}+\left(1+a_i\right)^p*-(1-p)*\left(1-a_i\right)^{-p}\\ =\left(1+a_i\right)^{p-1}*\left(1-a_i\right)^{-p}*\left(2*p-1-a_i\right) \end{align} $$Because $p<1$, we need not worry about $0^0$, this has roots at $a_i\in\left\{-1,1,2*p-1\right\}$. The $-1$ root is impossible, and both can be disqualified by the fact that they're also the roots of the equation we're maximising (which, note, for $p=\frac{1}{2}$ is the familiar $\sqrt{1-x^2}$ semicircle graph), so their $\log$s cannot be evaluated, so to optimise it, you should always bet proportion $2*p-1$ of your current money.