Math for fun: Solve for $x$ in $(-5)^x=3^{2x+3}$

Question

So I was bored, and decided to do some math for fun. After a while, I was bored because I was finding that the questions I was making for myself were a little too easy. So, I decided to create this equation:$$(-5)^x=3^{2x+3}$$where I need to solve for $x$. Here is my attempt at doing so:$$(-5)^x=3^{2x+3}$$Now, before I take the natural logarithm, here is how I am defining the complex logarithm:

Say there is a number $w$ where $e^w=z$. If $z$ is given in polar form $z=re^{i\theta}$, then all of the complex logarithms defined for $z$ is given in the form $\ln(r)+i(\theta+2\pi k)$, which are spaced evenly on a line in the complex numbers $\mathbb{C}$. Now to take the natural logarithm:$$(-5)^x=3^{2x+3}$$$$\implies x\ln(-5)=(2x+3)\ln(3)$$$$\implies i\pi x+x\ln(5)=2x\ln(3)+3\ln(3)$$Although really, this should be written as $$\implies\ln(-1\cdot5)=2x\ln(3)+3\ln(3)$$$$\implies\ln(e^{i\pi+2i\pi k}\cdot5)=2x\ln(3)+3\ln(3)$$$$\ln(5)+i\pi+2i\pi k=2x\ln(3)+3\ln(3)$$$$\because-1=e^{i\pi+2i\pi k}$$However, here is how this would be done continued from what I have:$$i\pi x+x\ln(5)=2x\ln(3)+3\ln(3)$$$$2x\ln(3)+3\ln(3)-i\pi x-x\ln(5)=0$$$$x(2\ln(3)-i\pi-\ln(5))+3\ln(3)=0$$$$\implies x\left(\ln\left(\dfrac95\right)-i\pi\right)=-3\ln(3)$$$$\because\ln(a)-\ln(b)\gets\ln\left(\dfrac ab\right)$$$$\implies x^2\left(\dfrac{\ln\left(\dfrac95\right)}x\right)-i\pi x+3\ln(3)=0$$However, it seemed that at this point, I had accidently made a math equation that is most likely unsolvable unless plugged into a computer application that can do math. So I plugged it into Wolfram Alpha and got this:

However, after some thought, I realized that there was something that I could have done, and decided to pretend like I hadn't done that. Here is what I then did: I went back to the $x\left(\ln\left(\dfrac95\right)\right)-i\pi=3\ln(3)$ and set $\ln\left(\dfrac95\right)-i\pi=a$. Then,$$ax=-3\ln(3)$$$$x=\dfrac{-3\ln(3)}a$$$$x=\dfrac{3\ln(3)}{2\ln(3)-\ln(5)-i\pi-2i\pi k},k\in\mathbb{Z}$$Which you might be asking, "Where did the $-2i\pi k$ come from? That wasn't there before!" Well here's how:

It's because I never, in any step, cancelled out the $i\pi$ that I was using. I was simply just going to plug back in the $2i\pi k$ when I found my solution, and it is valid because even though $-1$ is most commonly written in polar form as $e^{i\pi}$, it is also written as $e^{i\pi +2i\pi k},k\in\mathbb{Z}$ as a more definite definition of $-1$ in polar form! Then, set $2\ln(3)-\ln(5)=a$ and $\pi-2\pi k=b$ to put the denominator into $a\pm bi$ form, and then we can multiply by the conjugate:$$x=\dfrac{3\ln(3)}{a-bi}$$$$x=\dfrac{(3\ln(3))(a+bi)}{\require{cancel}(a\cancel{-}+b\cancel{i})(a+b\cancel{i})}$$$$x=\dfrac{(3\ln(3))(2\ln(3)-\ln(5)+i\pi+2i\pi k)}{(2\ln(3)-\ln(5))^2+\pi^2+4\pi^2k+4\pi^2k^2},k\in\mathbb{Z}$$

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My question

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Is the solution that I have arrived at correct, or what could I do to attain the correct solution/attain it more quickly?

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Mistakes I might have made

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1. Defining the complex logarithm
2. Logarithm simplification

$\small\text{yeah this might have been a little too difficult in all honesty}$

Answer 1

$ (-5)^x=3^{2x+3}$

$ \Rightarrow log(-5) * x = log(3) * (2x + 3) $

$ \Rightarrow \frac{log(-5)}{log(3)} *x = 2x + 3$

$ \Rightarrow \frac{log(-5)}{log(3)} *x - 2x = 3$

$ \Rightarrow x\left(\frac{log(-5)}{log(3)} - 2\right) = 3$

$ \Rightarrow x = \frac {3} {\left(\frac{log(-5)}{log(3)} - 2\right)}$

Answer 2

4 months later, I think I have a solution. So first we do$$\begin{align}(-5)^x=3^{2x+3}\implies&\,x\ln(-5)=2x\ln(3)+3\ln(3)\\\implies&\,x\ln(-5)-2x\ln(3)=3\ln(3)\\\implies&\,x(\ln(-5)-2\ln(3))=3\ln(3)\\\implies&\,x=\dfrac{3\ln(3)}{\ln(-5)-2\ln(3)}\\=&\,\dfrac{3\ln(3)}{\ln(-1)+\ln(5)-2\ln(3)}\\=&\,\dfrac{3\ln(3)(a-bi)}{(a+bi)(a-bi)}\\=&\,\dfrac{3\ln(3)(\ln(5)-2\ln(3)-\ln(-1))}{\ln^2(5)-4\ln(3)\ln(5)+4\ln^2(3)+\pi^2}\\=&\,\dfrac{3\ln(3)\ln(5)-6\ln^2(3)-3\ln(3)(i\pi+2i\pi k)}{\ln^2(5)-4\ln(3)\ln(5)+4\ln^2(3)+\pi^2}\\\approx&\,-0.19804726-1.05851979i\end{align}$$