Here is the context of the question: $X\sim \text{Gamma}(10,2)$
I want to derive the tilted density of $f$, where $f$ is the pdf of $X$. The tilted density is defined as $$\frac{e^{tx} \cdot f(x)}{M(t)}$$. The hint given to us is that we don't have to work out $M(t)$, I don't get how to make use of the hint to do this. Any help would be deeply appreciated :)
The density function $f$ of $X\sim \text{Gamma}(\alpha, \beta)$ is equal to $$ f(x)=\begin{cases}\dfrac{\beta^{\alpha}x^{\alpha-1}e^{-x\beta}}{\Gamma(\alpha)} & \text{if } x>0\\ 0 & \text{if } x\leq 0 \end{cases} $$ If your definition of Gamma distribution has another parametrization, you can modify the next equalities in accordance to it.
Consider only the part for $x>0$: the tilted density is $$\tag{1}\label{1} e^{tx}\dfrac{\beta^{\alpha}x^{\alpha-1}e^{-x\beta}}{\Gamma(\alpha)}\cdot\frac{1}{M(t)}=\dfrac{\beta^{\alpha}x^{\alpha-1}e^{-x(\beta-t)}}{\Gamma(\alpha)}\cdot\frac{1}{M(t)}. $$ This will be a density function on $(0,+\infty)$ for the case when $t<\beta$ only. For that case it should be a density of Gamma distribution with parameters $\alpha$ and $\beta-t$. So, it should looks like $$\tag{2}\label{2} f_{\text{tilted}}(x)= \dfrac{(\beta-t)^{\alpha}x^{\alpha-1}e^{-x(\beta-t)}}{\Gamma(\alpha)}, \quad x>0. $$ Note that you can equate both right hand sides of (\ref{1}) and (\ref{2}) and calculate $M(t)$ if you need.
Note finally that $e^{tx}\cdot f(x)/M(t)$ cannot be a density function in the case $t \geq \beta$ since the product becomes non-integrable for this case.