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Bonne chance!!

@@drpeyam oyo merci :D

Thank youuu :)

Hahaha

The power rule for logs is only valid with positive arguments so although this gives you the correct answer in this case it's not a valid approach in general.

Exactly what Andrey said! Here the rule is valid because a is a positive real number, but in general the rule is very dangerous

The π/ln2 term came from e^πi. That is - 1. For +1 the term is e^(0i+2πiM), so right between all those (1+2M)π lines

Agreed

Thank you!!

I’d review real analysis first

@@drpeyam Thanks a lot Professor!

There are some Hopital's videos on my channel discussing this, check them out

It works for f(x)/g(x) as long as the limit is an indeterminate of the form 0/0 or ±∞/±∞, and as long as g'(x) is non-zero in an open interval around a finite limit for x (except at the limit itself) or for sufficiently large x (for limit at ∞), or for sufficiently large negative x (for limit at -∞).

Thank youuuuu

Remember than ln can be negative, like ln(1/2) = -ln(2)

Doesn’t matter, here in the end it’s real

yeah i think so

We want to solve for complex z in 2^z=re^ipi. e is the base of natural logarithms so 2=e^ln2, and so 2^z=e^zln2, which allows the most direct comparison between the two sides of the original equation

@@hywelgriffiths5747 Yeah cheers I see it now. I think I'm just lazy sometimes in thinking about the whys of these steps.

Hahaha

But it is when it’s real

@@drpeyam Yes, but why then would you assume x is complex? The viewers are supposed to understand by f(x) = 2^x that you mean that the codomain is real, and that the domain is some subset of the complex numbers such that the function has real outputs? You've deliberately chosen to be ambiguous: why didn't you just write for what values of z does 2^z have an argument of pi. Say what you mean - don't just create click bait.

log_2(-a) with a>0 doesn't exists

@@finmat95 depends if we are in ℝ or ℂ

@@plutothetutor1660 Nope

@@finmat95 It does if x∈ℂ

To an engineer 1.1 will do for 1 but bloody mathematicians are purists and always insust on exact solutions. Which is not really what real life is like.

Presuming that your sin function accepts/interpretes its input in radians, the answer is yes.

@@Apollorion yes it accepts radian and input is in radian notation.

@Apollorion Can you elaborate? This is interesting

@@drpeyam he is saying that it is indeed positive or negative, i.e. not zero, since no integer except zero is a multiple of π. It would actually still be true if the argument were to be in degrees, as the given argument is an odd integer. Not deep at all, I'm afraid.

@Michael Rothwell Awwww I thought he meant a definite answer, like it’s definitely positive or definitely negative

I like that!!

Thanks so much!! :)

Thank you!!!

Wait what?

2^x is defined to be the positive root. If it had two values, 2^x would not be a function.

​@@drpeyam Yeah, it's something I learnt in middle school, long before complex numbers or precalculus. Did they force you to forget all that when you did your PhD?!!! 😆 Let me give you another example. This time let me use 9 as the base instead of 2, simply to avoid confusion with the radical sign, nothing more: _Can 9^x be negative?_ Let's see, (-3)² = 9, so 9^(1/2) = -3 is indeed a solution to 9^x < 0.

That is nonsense

​@@ramoncatalangonzalez8301 Dude, it's kinda late for jokes-today is the 3rd of April in my city; give or take a day in yours 🤣. In your original, unedited comment you said two things: (1) _"2^x is defined to be the positive root."_ Correction: Nope, 2^x is a power. Depending on x, the result can be anything-positive, negative, real, imaginary, complex, matrix etc. You're confusing it with the radical sign "√", which stands for the *principal* square root. Note: 'Principal' does not necessarily mean 'positive.' In fact, √x is positive if and only if x is positive. So √2 ≈ 1.4142... is positive. Otherwise √x is zero, imaginary, or complex. (2) _If it had two values, 2^x would not be a function._ Correction: Actually, 2^x is single-valued *only when x is an integer, otherwise it is **_multi-valued._* In particular, 2^(1/2) has two values, √2 ≈ 1.4142... and -√2 ≈ -1.4142... Did you even watch this video-or any video by Dr. Peyam where he tacks on 2πmi to the answers? Those are all multi-valued 'functions.' In fact, most 'functions' of complex arguments are multi-valued, even if their real counterparts are single-valued.

Danke 😁