Suppose $S$ and $T$ are self-adjoint operators in $B(H)$. we can use the following continuous functional calculus :
$C(\sigma(S)) \to C^*(S)$, $f\mapsto f(S)$ $C(\sigma(T)) \to C^*(T)$, $g\mapsto g(T)$ where $f(z)=e^z,g(z)=e^z$,we get $e^S$ and $e^T$.
How to use the continous functional calculus to show that if $S\neq T$,then $e^S\neq e^T$.Moreover,if we have the map $\psi$ from $B(H)_{sa}\to B(H)_{sa}$ such that $\psi(L)=e^{L}$,where $B(H)_{sa}$ is the set of self-adjoint elements in $B(H)$,how to deduce that $\psi$ is a bijective map?
If we let $S,T\geq 0$,we can get $log(S)$ and $log(T)$,if $S\neq T,$ how to conclude that $log(S)\neq log(T)$?
Your map is never bijective, since $e^S$ is always positive for $S$ selfadjoint.
Because $S$ is selfadjoint, $\sigma(S)\subset\mathbb R$. So $\sigma(e^S)\subset(0,\infty)$ and there is no issue applying the log function: over $\mathbb R$ we have ${\log e^t}=t$. So $$ \log(e^S)=S. $$
The above also shows that $S\neq T\implies\log S\ne\log T$; because the implication is the same as $\log S=\log T\implies S=T$, which is easily proven by taking the exponential.