## Calculus (3rd Edition)

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