# Logarithms of Negative Numbers

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?

• If I formatted something wrong or if you need clarification, please let me know.
– user192061
Jan 21 '15 at 20:13
• "Given the problem log2 (x) + log2 (x+2) = 3. The given solution excluded -4 as extraneous. I plugged it in, getting log2 (8) + 2[log2 (-1)] = 3" I'm not sure what you did here. Jan 21 '15 at 20:15