Show that $P(X>t)\leq \frac{E(e^{cX})}{e^{ct}}$
The claim isn't true for $c<0$. For example, let $X$ be uniform (0,1), $c=-10$ and $t=1/c$.
For $c=0$, the claim is trivial. For $c>0$, we have: $$ \Big[X>t\iff\exp(cX)>\exp(ct)\Big]\implies\Pr(X>t)=\Pr[\exp(cX)>\exp(ct)] $$ Now use Markov's inequality with $Y=\exp(cX)$.