Rewrite this as $\DeclareMathOperator{\tr}{tr} \ln \det \gamma = \tr \ln \gamma$. Then, we can switch to an eigenbasis of the matrix $\gamma$ and
$\ln \det \gamma = \sum_i \ln \lambda_i$
where $\lambda_i$ are the eigenvalues of $\gamma$. Since the eigenvalues of $\ln \gamma$ are $\ln \lambda_i$ we have
$\sum_i \ln \lambda_i = \tr \ln \gamma$
and thus
$\ln \det \gamma = \tr \ln \gamma$.