In Algebra II, I learned that you cannot take the logarithm of a negative number. However, when visiting the topic again, I realized that the identity $e^{i \theta} = \cos{\theta} + i\sin{\theta}$ gives a way to solve for this.
Given the problem $\log_2 {x} + \log_2 (x+2) = 3.$ The given solution excluded -4 as extraneous. I plugged it in, getting $$\log_2 {8} + 2(\log_2 (-1)) = 3,$$ which resolves into $$2(\log_2 (-1)) = 0,$$ $$\frac{2(\log (-1)}{\log (2)} = 0.$$ I then used $e^{i\pi} = -1$, which follows that $\log (-1) = \frac{i\pi}{\ln (10)}$. When plugging back in the value of $\log (-1)$, I saw that everything had a real value except $i$, which then means $i = 0$. What am I doing wrong?
In the complex numbers, you can take the logarithm of negative numbers as you are thinking. Unfortunately, the answer is not unique because of the periodicity of $\sin$ and $\cos$. From $e^{i\theta}=\cos \theta + i\sin \theta$ you also get $e^{i(\theta+2\pi)}=\cos \theta + i\sin \theta$, so you can add any integral multiple of $2\pi i$ to the log and get another value. You can restrict the values of the integer part of the log function to an interval of length $2 \pi i$ (say to $[-\pi i,\pi i)$ )to make it single valued, similar to what we do with $\arcsin$ and $\arccos$