Let \(X\) be a random variable and a, b be any constants. Then
$$M_{aX+b}(t)=\mathrm{e}^{tb}M_{X}(at)$$
Proof
\begin{eqnarray*}M_{aX+b}(t)&=&E[\mathrm{e}^{(aX+b)t}]\\&=&E[\mathrm{e}^{atX}\mathrm{e}^{bt}]\\&=&\mathrm{e}^{bt}E[\mathrm{e}^{atX}]\ \ \ (\text{ ∵ } \mathrm{e}^{bt} \text{ is a constant.})\\&=&\mathrm{e}^{tb}M_{X}(at)\end{eqnarray*}