okay let's do some math for fun here i have a very interesting question for you guys
is it possible to have 1 to some power and the result is equal to 2 (i.e. can we solve the exponential equation 1^x=2?)
what do you guys think well let me tell you if you enter this on WolframAlpha you will see
right here it shows no solutions it doesn't even give you any complex solution for this
that's really bizarre right so maybe okay that's the end of the video but no of course not
let's still try to solve it and let's try to investigate if we can actually get some
answer maybe from the complex world that the wolf alpha couldn't so let's just give you a try yeah
and the result is going to be pretty interesting so you just have a look
anyway i'm going to write this down again 1 to the x power and this is equal to 2
well by the look of it it's like an exponential function because the x is in the exponent
so i would like to take logarithm and then the log question that i'm going to use is
just log base 1 it's the best one that we can use right here because they cancel out
all right anyway though the left-hand side we will just get x by itself and then on
the right-hand side i will use the change of base boom block which will be ln 2 over
ln one and now you might see it yes ln one is zero so we have ln2 over zero that's not defined
so maybe i just really cannot get a solution for this
but you know what don't give up all we have to do is bring the ones to the complex world
and maybe make it into its complex form maybe you can continue much better so have a look
right here let me do it on the side for you guys one on the complex plane is going to be right here
and i want to change that to its complex one we need two things the first thing is the distance
from the original here which is just equal to one and then the next thing that we need is
to erase the equal sign better and then the third thing that we need is the angle from here to here
which is just going to be zero so let me just indicate that theta is equal to zero all right
cool now here we have ln2 over here we have ln one put that into this polar form we need r e i theta
r is one so we just have e raised to the i and the theta is equal to zero
if i cancel the ln and e i get i zero i'm still dividing by zero
but you know what this zero is not the only angle that will end up right here in this direction
because if you really want to do some rotation of course you can starting from here you can rotate
260 degrees but we all don't know so say that as 2 pi and if you were like you can do the reverse
negative 2 pi and if you really like you can do two loops which is
4 pi etc etc so the truth is you have infinitely many angles can enter in this direction and we
really have to write down the whole thing so the whole thing is just actually multiple of
two pi so we'll just plus two pi n where n is just an integer it can be positive or negative
whole number so we will actually have to attach that right here as well so i will just add 2 pi
n right that's actually our data we have to look at it's all this value because we have multi value
for this situation now we can cancel this out and we don't really have to have a zero on the bottom
anymore and then what we have is just going to be ln 2 on the top and then we have i and just
that so i'll just write it as i times 2 pi 10. but you know me i don't like to be on the bottom right
to be on the top so let's multiply the bottom and the top by i so that this right here will
be negative one so only o i will write this as negative one on the top with the i and then ln2
over two pi and for two and pi depending on how you multiply it and before i box this you
see the units on the bottom right so you have to make sure that n cannot be equal to zero
and cannot be equal to zero just like that now how many solutions do we actually have
well it's kind of like infinitely many but each every time you put an answer let's say you choose
n to be one if you put it back let me just show you how it looks like yeah so let's say we take
n to be one so we just have this guy a so what we get is just one
to the power of negative i ln2 over two pi
will this give us two the answer to this is yes but not
all the time and the reason is because if you really want to do the complex analysis when you
do one to this power you have to write one in terms it's polar form and then you have
infinitely many answers for that right and you have to work out some exponents business
yeah and you have infinitely many answers and uh one of the answers will be two
it's a very weird computation right so hopefully you guys are okay with that but before we end
i wanted to talk about you see 1 to the x power is equal to 2 and we'll find alpha
it does not give you any solution however if you look at 1 to the x power is equal to 2
as 1 equals 2 to the 1 over x power then you just take the x on both sides
if you enter this on the wall from alpha it actually works um pretty interesting i think it's
just because how they compute one to some power is just always one because x is like a real number so
that's pretty much why i think i really don't know about once you enter this you will see that it
will actually give you an answer like that right of course and cannot be good certainly will keep
that answer really cool right so hopefully you guys all like this video and as always that's it