# Exponent laws with Negative bases

$-49 =7^x$ is the question. Here I m supposed to solve for what power of $7$ will give me $-49$. Or in other words, I have to solve for $x$. This looks fairly simply when thinking about the exponent rules for it looks as if you could make the $-49$ become $7$ to the power of $2$ so that the bases are common. However, it is a negative so I can't do this. I am unsure how to approach this simply because of the negative on the left side of the expression. Please help me out.

$7^x = e^{x \ln 7}$.
Now you want $$e^{x \ln 7} = -49 = e^{\Re (x \ln 7)} e^{i \Im (x \ln 7)} \equiv e^R e^{i \theta}$$
where $R \equiv \Re(x) \ln 7$ and $\theta \equiv \Im(x) \ln 7$.
Since the final form of the product can be expressed as the product of a magnitude and direction in the complex plane, and we know the magnitude is 49 and direction is left, we have \begin{align} e^R &= 49 \implies \Re(x) = \frac{\ln 49}{\ln 7} = 2 \\ e^{i \theta} &= -1 \implies \Im(x) = \frac{\pi + 2 \pi n}{\ln 7} \end{align} where $n$ is any integer.
Thus $x = 2 + i \frac{2 \pi (n+1/2)}{\ln 7}$ for any integer $n$.