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- How do I square a logarithm? - Mathematics Stack Exchange
$\log_2 (3) \approx 1 58496$ as you can easily verify $ (\log_2 (3))^2 \approx (1 58496)^2 \approx 2 51211$ $2 \log_2 (3) \approx 2 \cdot 1 58496 \approx 3 16992$ $2^ {\log_2 (3)} = 3$ Do any of those appear to be equal? (Whenever you are wondering whether some general algebraic relationship holds, it's a good idea to first try some simple numerical examples to see if it is even possible
- Mathematics Stack Exchange
Q A for people studying math at any level and professionals in related fields
- Legendres three-square theorem - Mathematics Stack Exchange
Legendre's three-square theorem Ask Question Asked 5 years, 2 months ago Modified 1 year, 7 months ago
- What is $\sqrt {i}$? - Mathematics Stack Exchange
The square root of i is (1 + i) sqrt (2) [Try it out my multiplying it by itself ] It has no special notation beyond other complex numbers; in my discipline, at least, it comes up about half as often as the square root of 2 does --- that is, it isn't rare, but it arises only because of our prejudice for things which can be expressed using small integers
- algebra precalculus - How to square both the sides of an equation . . .
I understand that you can't really square on both the sides like I did in the first step, however, if this is not the way to do it, then how can you really solve an equation like this one (in which there's a square root on the LHS) without substitution?
- Why sqrt(4) isnt equall to-2? - Mathematics Stack Exchange
If you want the negative square root, that would be $-\sqrt {4} = -2$ Both $-2$ and $2$ are square roots of $4$, but the notation $\sqrt {4}$ corresponds to only the positive square root
- Finding Two Distinct Square Roots of I [closed]
Finding Two Distinct Square Roots of I [closed] Ask Question Asked 1 year, 9 months ago Modified 1 year, 9 months ago
- When does $\sqrt {a b} = \sqrt {a} \sqrt {b}$? - Mathematics Stack Exchange
Now, since we are working in polar form, we can evaluate the square roots consistently, arriving at $$ 1=e^ {-\pi i 2}\times e^ {\pi i 2} = -i\times i = 1 $$ Essentially, the problem lies in the "branch cut" that occurs with the square root operation - you must be careful with the evaluation
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