<s>[INST] 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.
 [/INST] $$(1+2i)^{(2x+6)}=(-11-2i)^{(x+1)}$$ is equivalent to $$(1+2i)^{(2x+6)}=(1+2i)^{3(x+1)}$$ or, assuming $(1+2i)^{(2x+3)}\neq 0,$ $$(1+2i)^{3}=(1+2i)^{x}.$$
One solution is $x=3.$
I do not think that other solutions would be trivial to find.
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