Answer
$$\langle \ln 6,e+\pi \rangle$$
Work Step by Step
$$
\left\langle \ln 2, e\right\rangle+\left\langle \ln3, \pi\right\rangle=\left\langle \ln 2+\ln3, e+\pi\right\rangle
$$
Recall the log rule:
$\log(a\times b)=\log(a)+\log(b)$
$$=\langle \ln(2\times 3),e+\pi \rangle \\
=\langle \ln(6),e+\pi \rangle$$
