Negative Logarithms?

$\begin{aligned} \log_2 (-1024) & = \frac {\ln (-2^{10})}{\ln 2} & \small \blue{\text{Convert base 2 to base }e} \\ & = \frac {\ln (2^{10}\blue{(-1)})}{\ln 2} & \small \blue{\text{By Euler's formula: }e^{i\theta} = \cos \theta + i \sin \theta} \\ & = \frac {\ln (2^{10}\blue{e^{\pm i\pi}})}{\ln 2} & \small \blue{\text{Note that }\log(ab) = \log a + \log b} \\ & = \frac {10 \ln 2 + \ln e^{\pm i \pi}}{\ln 2} \\ & = 10 \pm \frac \pi{\ln 2}i \end{aligned}$

Then $\left(10+\dfrac \pi{\ln 2}i \right)\left(10-\dfrac \pi{\ln 2}i \right) = 10^2 + \left(\dfrac \pi{\ln 2} \right)^2 \approx \boxed{120.542}$ .

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Reference: Euler's formula

Euler's formula tells us that $e^{(2k - 1)\pi i} = -1$ for any integer $k$ . From this, if we take the natural logarithm of both sides, we get $\ln(-1) = (2k - 1)\pi i$ . We can get the general form $\ln(-n)$ using the property $\ln(ab) = \ln(a) + \ln(b)$ . Since $\ln(-n) = \ln(n * -1) = \ln(n) + \ln(-1)$ , we conclude that $\ln(-n) = \ln(n) + (2k - 1)\pi i$ . Dividing all sides by $\ln(2)$ gives us the formula $\log_2(-n) = \log_2(n) + \frac{(2k - 1)\pi}{\ln(2)} i$ .

Thus, the two solutions are $10 + \frac{\pi}{\ln(2)}i$ and $10 - \frac{\pi}{\ln(2)}i$ and their product is $(10 + \frac{\pi}{\ln(2)}i)(10 - \frac{\pi}{\ln(2)}i) = 100 + \frac{\pi^2}{\ln^2(2)}= \boxed{120.542}$

$\log_2 (-1024) = \log_2 (1024) + \log_2 (-1)$ $= 10+ \frac{\log(-1)}{\log(2)} = 10+\frac{\log(e^{±πi})}{\log(2)}$ Two solutions are $y_1 =10 + \frac{πi}{\log(2)} ,y_2 = 10-\frac{πi}{\log(2)}$ $y_1 .y_2 = 10^2 +\frac{π^2}{\log^2( 2)} = \color{#20A900} \boxed{120.5422}$