Suppose $\{B_t,t\ge0\}$ be a standard brownian motion and suppose $0\le u\le s
Attempts: $E(e^{B(t)}|e^{B(u)},0\le u\le s )=E(e^{B(t-u)+B(u)}|e^{B(u)})=E(e^{B(t-u)}|e^{B(u)})E(e^{B(u)}|e^{B(u)})$ but then not sure how to proceed.
Suppose $\{B_t,t\ge0\}$ be a standard brownian motion and suppose $0\le u\le s
Attempts: $E(e^{B(t)}|e^{B(u)},0\le u\le s )=E(e^{B(t-u)+B(u)}|e^{B(u)})=E(e^{B(t-u)}|e^{B(u)})E(e^{B(u)}|e^{B(u)})$ but then not sure how to proceed.