Solve for $x$, $(1+2i)^{(2x+6)}=(-11-2i)^{(x+1)} $ $$(1+2i)^{(2x+6)}=(-11-2i)^{(x+1)} $$
Solve for $x$
My Work:
Take ln at the both sides:
$$(2x+6)\ln(1+2i)=(x+1)\ln(-11-2i)$$
and$$\ln(1+2i)=\ln(\sqrt{5})+i\arctan(2)$$
$$\ln(-11-2i)=\ln(5\sqrt{5})+i\left[\arctan\left(\frac{2}{11}\right)+\pi\right]\\=\ln(5)+\ln(\sqrt{5})+i\left[\arctan\left(\frac{2}{11}\right)+\pi\right]$$
And things get dirty here because of $\arctan$ and $\ln$ ...
But
$$(2x+6)\ln(1+2i)=(x+1)\ln(-11-2i)$$ from here
$$2x+6=(x+1)\dfrac{\ln(-11-2i)}{\ln(1+2i)}$$
then
$$x\left(2-\dfrac{\ln(-11-2i)}{\ln(1+2i)}\right)=-6+\dfrac{\ln(-11-2i)}{\ln(1+2i)}$$
then
$$x=\dfrac{-6+\dfrac{\ln(-11-2i)}{\ln(1+2i)}}{2-\dfrac{\ln(-11-2i)}{\ln(1+2i)}}$$
But specifically I cannot find x here.
Any elegant way to solve it, or not elegant but proper, any hint any help, would be appreciated. Thank you.

Hint:
Observe that $11^2+2^2=\cdots=5^3, 5=1^2+2^2$
which encouraged me  to find $$(1+2i)^3=1-12+(6-8)i$$
$$\implies(-11-2i)^{(x+1)}=((1+2i)^3)^{x+1}=?$$
So we have $$(1+2i)^{3(x+1)-(2x+6)}=1$$
Taking logarithm in both sides, $$(x-3)\ln(1+2i)=2n\pi i$$ where $n$ is any integer
Now $1+2i=\sqrt{1^2+2^2}\cdot e^{i\arctan2}$
$\implies\ln(1+2i)=\dfrac{\ln5}2+i\arctan2+2m\pi i$  where $m$ is any integer
See also: Find all real numbers $x$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac76$
