This snapshot is taken from Principles of Optics by Born and Wolf.
I'm not able to comprehend the equation $(3)$ above.
First they take the identity $$\textrm{curl}~{u\mathbf v} = u~\textrm{curl}~{ \mathbf v} + \left(\textrm{grad}~ u\right)~\boldsymbol\land~ \mathbf v,$$ where $u$ is a scalar field and $\mathbf v$ is a vector field.
Using it in $(2),$ we get $$\textrm{curl}~\left(\frac1\mu \textrm{curl}~\mathbf E\right) = \frac1\mu ~\textrm{curl}\left(\textrm{curl}~ \mathbf E\right) + \left(\textrm{grad}~\frac1\mu\right)~\boldsymbol\land~ \textrm{curl}~\mathbf E\,.$$
So, the equation $(3)$ would be $$\boldsymbol\nabla^2~\mathbf E - \frac{\varepsilon\mu}{c^2}~\ddot{\mathbf E}+ \left(\textrm{grad}~\frac1\mu\right)~\boldsymbol\land~ \textrm{curl}~\mathbf E - \textrm{grad}~\textrm{div}~\mathbf E\,. $$
But this is not the same as given in the snapshot above viz. $$\boldsymbol\nabla^2~\mathbf E - \frac{\varepsilon\mu}{c^2}~\ddot{\mathbf E}+ \left(\textrm{grad}~\color{red}{\log \mu}\right)~\boldsymbol\land~ \textrm{curl}~\mathbf E - \textrm{grad}~\textrm{div}~\mathbf E\,. $$
They have done same sort of things in the later equations too.
I couldn't get that.
The identity was talking about $u\mathbf v;$ the gradient term, as the writers wrote, was $\textrm{grad}~u\,.$ So, in this context, that would have meant $\left(\textrm{grad}~\frac1\mu\right)$ and not $\textrm{grad}~\log \mu$ as they wrote.
How did they get the $\log\mu$ term? Isn't it contradicting the identity?
The book is correct (after all it's the sixth or seventh edition).
Observe \begin{align} \operatorname{curl}\left(\frac{1}{\mu}\operatorname{curl}\mathbf{E} \right)=&\ \frac{1}{\mu}\operatorname{curl}\operatorname{curl}\mathbf{E}+\operatorname{grad}\left(\frac{1}{\mu}\right)\wedge\operatorname{curl}\mathbf{E}\\ =&\ \frac{1}{\mu}\operatorname{curl}\operatorname{curl}\mathbf{E}- \frac{1}{\mu}(\operatorname{grad}\log \mu)\ \wedge \operatorname{curl}\mathbf{E}\\ =&\ \frac{1}{\mu} \operatorname{grad}\operatorname{div}\mathbf{E}-\frac{1}{\mu}\nabla^2\mathbf{E}- \frac{1}{\mu}(\operatorname{grad}\log \mu)\ \wedge \operatorname{curl}\mathbf{E} \end{align} which means \begin{align} \operatorname{curl}\left(\frac{1}{\mu}\operatorname{curl}\mathbf{E} \right)+\frac{\varepsilon}{c^2}\mathbf{\ddot E} = \frac{1}{\mu} \operatorname{grad}\operatorname{div}\mathbf{E}-\frac{1}{\mu}\nabla^2\mathbf{E}- \frac{1}{\mu}(\operatorname{grad}\log \mu)\ \wedge \operatorname{curl}\mathbf{E}+\frac{\varepsilon}{c^2}\mathbf{\ddot E}=0. \end{align} Hence it follows \begin{align} \nabla^2\mathbf{E}-\operatorname{grad}\operatorname{div}\mathbf{E}+ (\operatorname{grad}\log \mu)\ \wedge \operatorname{curl}\mathbf{E}-\frac{\varepsilon\mu}{c^2}\mathbf{\ddot E}=0. \end{align}