Assume $p(x)= \log\mathcal{N}(x|\mu_p,\Sigma_p)$ and $q(x)= \log\mathcal{N}(x|\mu_q,\Sigma_q)$. Where $x\in R^D$ and $D$ is the dimension of $x$. So $x = [x_1,x_2, \ldots ,x_D]^T.$
Based on KL-divergence definition we have:
$$D_{KL}(p(x)||q(x))=\int\limits_{R^D} p(x)\log\dfrac{p(x)}{q(x)}=\int\limits_{R^D}p(x)\log (p(x)) - \int\limits_{R^D}p(x)\log(q(x))=E_p\{log(p(x))\} - E_p\{log(q(x))\}.$$
Before taking the first expectation, let's first determine the $\log(p(x))$:
$$log(p(x))=-\frac{D}{2}\log(2\pi)-0.5\log(|\Sigma_p|)-\sum\limits_{i=1}^Dlog(x_i) - 0.5(\log(x)-\mu_p)^T\Sigma_p^{-1}(\log(x)-\mu_p).$$
The important thing to remember is that if for example $z\sim\log\mathcal{N}(\mu,\Sigma)$ then $\log(z)\sim\mathcal{N}(\mu,\Sigma)$.
Now let's take the expectation of $\log(p(x))$:
$$E_p\{log(p(x))\}=-\frac{D}{2}\log(2\pi)-0.5\log(|\Sigma_p|)-\sum\limits_{i=1}^D\mu_p(i) - 0.5*trace(I_D).$$
Similarly for $E_p\{log(q(x))\}$:
$$E_p\{log(q(x))\}=-\frac{D}{2}\log(2\pi)-0.5\log(|\Sigma_q|)-\sum\limits_{i=1}^D\mu_p(i)-0.5*trace(\Sigma_p\Sigma_q^{-1})-0.5(\mu_p-\mu_q)^T\Sigma_q^{-1}(\mu_p-\mu_q).$$
So for $D_{KL}(p||q)$ :
$$D_{KL}(p||q)=0.5\log(\dfrac{|\Sigma_q|}{|\Sigma_p}|)-D/2 + 0.5 trace(\Sigma_p\Sigma_q^{-1})+0.5(\mu_p-\mu_q)^T\Sigma_q^{-1}(\mu_p-\mu_q).$$
you can see that the KL divergence between two lognormals is equal to the Kl divergence between two normal distributions with the same parameter $\mu$ and $\Sigma$.