What is the distribution of sum of $\ln(X_i)$ if $X_1, ..., X_n$ are from Gamma distribution?

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Let $X = (X_1, ... X_n)$ where $X_i$ are iid from distribution gamma with parameter $p$ in the following form: $$f_p(x) = \frac{\lambda^p}{\Gamma(p)}x^{p-1}\exp[-\lambda x]$$ for $x > 0$ and $0$ elsewhere. I learned that $T(X) = \sum{\ln(X_i)}$ is a sufficient statistic of $X$. In the next part of my task I have to assume that $P[T(X) > c] = \alpha$ for some $\alpha > 0$ and $c > 0$ and I think that I need to know $T(X)$ distribution.

Question: How can I know what distribution does $T(X)$ have?

Edit: Clarification of context

It's a part of exercise where I need to find UMP test with significance level $\alpha$ for testing $H_0: p = 2$ vs $H_1: p > 2$ using Karlin-Rubin theorem. In order to apply it I need to know that joint distribution of $f(X)$ has monotone likelihood ratio - it can be done by showing that $f(X)$ is from exponential family with term $T(X)$ being a sufficient statistic. Then (by K-R theorem) critical region is $C = \{T(X) > c\}$.

I am asked to find $T(X)$ and say something about $k$. The answer states, that $T(X)$ is $\sum{\ln(X_i)}$ and that $k$ depends on $\lambda$ and $\alpha$ - but I can't show why and I am also not sure why it isn't dependent on $p_0$. I thought that it can be shown from definition of significance level: $\alpha = P[T(X) > k]$ but I don't know how to compute that probability (as I don't know $T(X)$ distribution.

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Assuming that $\lambda$ is a known parameter, $T=\Sigma_i \log x_i$ is sufficient for $p$.

Now, reading your post, it is not clear what your are specifically asked to answer.

Just to simplify calculations and without loss of generality we can set $\lambda=1$ and, if necessary, the distribution of

$$Y=\log X$$

is easy to be derived, being

$$f_Y(y)=f_X[g^{-1}(y)]\left|\frac{d}{dy}g^{-1}(y) \right|=\frac{1}{\Gamma(p)}\exp\{yp-e^{y}\}$$

$y\in\mathbb{R}$

Now you can understand that the calculation of the distribution of $T=\Sigma_i Y_i$ does not look like easy...

Perhaps they asked you a different question?