Yep, that's what I came up with. For (A), x = ln(2) which is less than 1. For (B), we're taking the ln of a number less than 1, so x is negative. That means for (c) we need to ln a negative number.
Ever since I watched your channel, I always box in my answers. It's been super helpful for keeping my math work organised. Every time I solve any math problem, that box reminds me of your amazing work. And you even inspired a video of mine, so thank you so much! ✌️
Before watching: Option C Solution to the first one is ln(2), 0.693 approx Solution to the second one is ln(ln(2)), which is negative Solution to the third one is ln(ln(ln(2))) which is imaginary as input to the outermost ln function is negative.
c(x) = e^e^e^x is a strictly increasing function as x increases. If you look at the limit x -> -inf you can see, that e^e^e^x approaches e^e^0 = e^1 = e > 2 Therefore the domain of c doesn't contain any numbers smaller than or equal to e, which obviously includes 2, meaning: The equation has no real solution.
Can you make a Digital SAT Prep series? I wish this guy was my math teacher. Everything you do is amazing and I love watching your videos and trying complex problems beyond my understanding. Your explanations are so clear and concise. It doesn’t hurt to know more math.
My first thought before watching is that, we know e^x is continuously and increasing from x=-∞ to x=∞ Putting x=-∞ into e^x gives 0. Putting x=1 into e^x gives e. Therefore, e^-∞ < 2 < e^1 and as e^x is continuous there must be some real x which gives e^x = 2. Similarly doing this for the second option we have the function e^e^x again this is continuous from x=-∞ to x=∞ and increasing Putting x=-∞ gives 1 Putting x=0 gives e. Therefore, again we have e^e^-∞ < 2 < e^e^0 Due to continuity we know that some real x fulfils e^e^x = 2 Now the 3rd option, we basically do the same thing we now the function continuous and increasing across the whole x-axis. Then put values Putting x=-∞ gives e. Putting x=0 gives e^e. So we see that the minimum value for this function is e and e > 2 therefore there exists no real x which satisfies this equation.
Hi! I want to ask you if you can find these limits, and make a video where you show the answer, here are the limits: let n be a natural number, calculate these limits; 1. lim x->0 ((1-cos(x)cos(2x)...cos(nx))/x^2) 2. lim x->pi/2n (sqrt(sin(2nx)/(1+cos(nx)))/(4n^2*x^2-pi^2)) pay attention to the parentheses so you won't calculate a different limit, thanks.
I know this request is for somebody else, but maybe this helps: 1. Just use de l'Hospital's rule twice; 2. Substitute t = x - pi/2n, the limit doesn't exist because up to asymptotic equivalence the expression is sqrt(-t / 1) / t. By the way, this means that it should have been a limit from the left since otherwise sqrt(-t) is undefined.
I haven't made a comment on any of your videos. But I love your math videos even though I am not quite at the level of math to understand some of these complex concepts. These are fun to watch when I am bored and make me passionate about math.
Complex numbers are not such big deal, just remember that every number x = a + b*i (if b=0, then the number is real). Moreover, sometimes the complex coordinates help to visualize (where x is the real part and y is the imaginary part)
For the complex solution if you want to find all solutions, you need to attach +2n(use different var names obviously)ipi after _every logarithm_ If you want to find just one solution on the other hand, you don't need any additional +2nipi
One calc equation : What's the irl use of finding the volume when we rotate f(x) about a line using the Disc method, the Washer method etc, when we can simply....... Take the object and submerge it just below a given level of water of volume *v1 cubed units* and see the new reading of reaching the new greater value of *v2 cubed units* , and get the volume of the object as *v2-v1 cubed units* ??? [Done by Archimedes' Principle]
Can you please find the indefinite integral of e^-u^t, where t is a constant? If indefinite is impossible, then what is the integral from 0 to ∞ of e^-u^t
If we search for ALL complex solutions of e^e^e^x = 2, shouldn't we already assume that e^e^x may already be complex? And if we only search for only SOME solution(s), we don't need the "2*n" but we can simply chose n=0.
I think the explanation of having to add +2nipi doesn't get at the mathematics behind it. When you apply the complex "logarithm" on both sides the implication from the latter to the former statement will always be true. And indees 2ipi=0 does imply e^i2pi=1. Similarly when finding the solution to an equation taking an arbitrary log will yield *a* solution, just not the solution. If you add the +2nipi, you can look at it like this in general: e^x = e^y ==> There exists n in N s.t. x +2nipi = y This works for the e^i2pi=1 as well here it implies there exists n in N s.t. 2ipi + 2nipi = 0 and indeed, it does with n=-1
My first thought is if the answer is A or B, then C also qualifies, dinde e raised to the power of a number not in R will also not be in R. Therefore the only answers if the question is honest is either C or D. On a standardized test, this is already enough that a random answer is better than no answer. Then e is around 2.7, so A will have a real solution for x that is less than 1 but greater than 0 because e>2. To get a value less than 1 of e^x to get a root for B, we need x to have a value less than 0. To get a value less than 0 for e^x to get a root for C, we need x to be a complex number. That is not in R, so C is the answer.
I would like to ask if it's possible to solve the following set of non-standard simultaneous equations in x, y, and z for some nice values of a, b, and c (or maybe just a generalised solution)... x+y+z = a xyz = b x^(y^z) = c
Hey! If @blackpenredpen sees this, I have always wondered, if you took the Monty hall problem but made both final doors have an active participant in front of each, making their own decision, do the odds change of winning for anybody? If not or if so, what justifies the changes or lack of change? It bugs me thinking about it.
The setup fundamentally does not work with two active players and three doors in total. Player A picks door 1 Player B picks door 2 Monty now deliberately opens a door that is unpicked and doesn't have the car. But he can only do that in case one of them has the car. So either he can't open a door *or* he can, but then no one can change. If you allow for, say, 100 doors instead. Player A picks door 1. Player B picks door 2. Monty now deliberately opens 97 unpicked doors that don't have the car. Now players A and B would both like to switch to the unpicked, unopened door. The car is with p=0.01 behind door 1, with p=0.01 behind door 2 and with p=0.98 behind the unpicked door.
Hello, I'm trying to calculate the quantity \int_{0}^{\frac{\pi}{2}} \exp(-y\theta) \cos^{x-1}(\theta) \, d\theta which is a kind of generalized Wallis integral, if anyone has an idea.
Can you do a video on how to find the inverse of (1+1/x)^x=y? I am really curios because i know it exists but I don't know how to get it even with using the Lambert function
Hello bprp! Can you solve this problem from russian SDA (School of Data Analysis)? Graph F(x)=lim as n goes to inf(nth root(1+x^n+(x/2)^n) it seems overcomplicated (btw WA cant solve it), but turns out its not, haha
Sir! I have a question for you (pls make a video on it if u like) we know that 1^x=1 for all real x; but I have accidentally proved x to be a rational and not real. pls find the fault in my proof. Here goes: first lest take a general case 1^x=a therefore: x=ln(a)/ln(1) = ln(a)/ln(e^i2nπ) = ln(a)/i2nπ = -i ln(a)/2nπ {n belongs to Z} now for 1^x=1 we sub in '1' for 'a' and we get : x = -i ln(1)/2nπ = -i (i2kπ)/2nπ = 2kπ/2nπ = k/n {k and n belong to Z} therefore x=k/n which is the definition of a rational number. Pls find a way to fix this!! It will be appreciated.🙏
C. The natural log of ln(2) is negative, so its natural log is not a real number. It's one of those times where it's important to actually compute the answers, rather than just finding an exact solution.
One might also say that this is "one of those times where it's important to actually compute the answer" before posting a comment! 😛 Because if you calculate ln(2) it turns out that this is *not* negative after all.
@@jensraab2902 you may say this is one of those times where it's important to actually read the comment before responding to the comment - he says that the natural log of ln(2) is negative - that's ln(ln(2))
Third one can't have a solution befause sol. for A is ln(2) which is positive but less than e the natural log of something less than e is negative → ln(ln(2)) exists but is negative ln(ln(ln(2))) is therefore undefined
What 'n' are you even talking about? EDIT: if we are talking about the solutions for e^e^e^x = 2, then yes, n can be any integer, and yes, there are infinitely many complex solutions there, none of which are real.
Couldn’t we say all the fuction are bijection and all have a limit lower than 2 in -infinity and all go to infinity in + infinity? So D would be the answer?
Your channel is so good I actually had to UNSUBSCRIBE since I keep getting distracted with how interesting these are and will spend hours losing myself working out the solutions.
Oh, I missed the “real” in the title. I was thinking, “well, the image of exp (as a function from C to C) is everything other than zero, so for all but finitely many values of y, there will exist a z such that e^e^e^…e^z = y , right?”
This reminds me of an old joke told by one of my professors. e^z assumes all values except 0. Thus e^e^z assumes all values except 0 and 1. But it is an entire function, so the Picard's theorem implies that e^e^z is constant. ;P
@@adayah2933 Is the hole that e^e^z actually does take on the value 1, because 2pi i (for example) is in the image of e^z ? (Uh.. specifically at, z= ln(2pi) + i 2pi/4 ) Nice joke
You forgot something. It's actually: e^e^e^x=2 e^e^x=ln2+(2n+1)πi e^x=ln(ln2+(2n+1)πi)+(2n+1)πi x=ln(ln(ln2+(2n+1)πi)+(2n+1)πi)+(2n+1)πi I'm too lazy to calculate the complex lns but you get the idea.
Sir please kindly make videos on algebraic geometry and algebraic topology kindly please see my comment and make videos on it ❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤😢😢😢😢 I have been watching you for a very long timw
Your third line does not make sense, because it is neither an equation nor an equality, so it cannot have a "solution." It is a different question whether it has or not has a real value.
Can e^e^x=1?
ua-cam.com/video/ckc9F0VjZ3k/v-deo.html
of course not, because
e^x cannot be 0.
you could approach zero as a limit, if you take
x →-∞ .
?
@@keescanalfp5143 however, we can have eˣ=2nπi for n∈ℤ, n≠0, so there are plenty of solutions.
@@keescanalfp5143 e^x cannot be 0, but it can be 2πi (and e^2πi = 1)
@@keescanalfp5143 Well, not for real numbers.
e^e^x=1, so x=ln(ln(1));
x=ln(0) which is undefined
keep recursively taking those ln()'s and eventually you end up taking up taking the natural log of a negative number, which isn't real.
Yep, that's what I came up with. For (A), x = ln(2) which is less than 1. For (B), we're taking the ln of a number less than 1, so x is negative. That means for (c) we need to ln a negative number.
But it feels sooo real. I can't explain why. There must be a pie in my eye in the sky.
Also if you approach 0 while taking ln, you know you have arrived.
For real x, e^x > 0
Therefore e^e^x > 1
Therefore e^e^e^x > e
2
how did you coclude 2e accordingly with your calculations?
@@arturjorge2816 The solution requires that 2>e, so because we already know that 2
@@kicorse Oh yeah, of course. I thoght you wrote 2
Same
2 = e = π = 3
Ever since I watched your channel, I always box in my answers. It's been super helpful for keeping my math work organised. Every time I solve any math problem, that box reminds me of your amazing work. And you even inspired a video of mine, so thank you so much! ✌️
Thank you! 😃
Before watching:
Option C
Solution to the first one is ln(2), 0.693 approx
Solution to the second one is ln(ln(2)), which is negative
Solution to the third one is ln(ln(ln(2))) which is imaginary as input to the outermost ln function is negative.
You got it! great job.
Yay I got it too without calculator
Nice, I did it the same way
Me too!
I invite u to listen to this recitation of the Noble Quran. ua-cam.com/video/IKZ57sdcFK4/v-deo.htmlsi=i8lD0MDkwVHgdQA3
c(x) = e^e^e^x is a strictly increasing function as x increases. If you look at the limit
x -> -inf you can see, that e^e^e^x approaches e^e^0 = e^1 = e > 2
Therefore the domain of c doesn't contain any numbers smaller than or equal to e, which obviously includes 2, meaning: The equation has no real solution.
7:06 if this is a way of advertising your shirt it's hilarious 😂😂
"Again, buy my shirt!" lol
proof buy my shirt
But imagine if it did have a solution
Was going to answer to this but just got the joke. Good one lol
@@RefreshingShamrock lmao🤣🤣🤣🤣 night reference
It have but its comple,
@@s-1i-youtube you didn't get the joke, by "imagine", he means that it has a complex solution
You never disappoint, sir.
Can you make a Digital SAT Prep series? I wish this guy was my math teacher. Everything you do is amazing and I love watching your videos and trying complex problems beyond my understanding. Your explanations are so clear and concise. It doesn’t hurt to know more math.
My first thought before watching is that,
we know e^x is continuously and increasing from x=-∞ to x=∞
Putting x=-∞ into e^x gives 0.
Putting x=1 into e^x gives e.
Therefore, e^-∞ < 2 < e^1 and as e^x is continuous there must be some real x which gives e^x = 2.
Similarly doing this for the second option we have the function e^e^x again this is continuous from x=-∞ to x=∞ and increasing
Putting x=-∞ gives 1
Putting x=0 gives e.
Therefore, again we have e^e^-∞ < 2 < e^e^0
Due to continuity we know that some real x fulfils e^e^x = 2
Now the 3rd option, we basically do the same thing we now the function continuous and increasing across the whole x-axis. Then put values
Putting x=-∞ gives e.
Putting x=0 gives e^e.
So we see that the minimum value for this function is e and e > 2 therefore there exists no real x which satisfies this equation.
Hi!
I want to ask you if you can find these limits, and make a video where you show the answer, here are the limits:
let n be a natural number, calculate these limits;
1. lim x->0 ((1-cos(x)cos(2x)...cos(nx))/x^2)
2. lim x->pi/2n (sqrt(sin(2nx)/(1+cos(nx)))/(4n^2*x^2-pi^2))
pay attention to the parentheses so you won't calculate a different limit, thanks.
Yes, i tried it, it's hard and i would like to see the solution 😊
I know this request is for somebody else, but maybe this helps:
1. Just use de l'Hospital's rule twice;
2. Substitute t = x - pi/2n, the limit doesn't exist because up to asymptotic equivalence the expression is sqrt(-t / 1) / t. By the way, this means that it should have been a limit from the left since otherwise sqrt(-t) is undefined.
I haven't made a comment on any of your videos. But I love your math videos even though I am not quite at the level of math to understand some of these complex concepts. These are fun to watch when I am bored and make me passionate about math.
Complex numbers are not such big deal, just remember that every number x = a + b*i (if b=0, then the number is real). Moreover, sometimes the complex coordinates help to visualize (where x is the real part and y is the imaginary part)
The graph of c seems to have a vertical asymptote very close to the y axis, which is fascinating.
e^x > 0, so e^e^x must be greater than e^0. Then e^e^x > 1. Finally, e^e^e^x > e^1 > 2
The easiest way to solve this
Upload more man 😢😢 I'm missing your vids already
Will do!
For the complex solution if you want to find all solutions, you need to attach +2n(use different var names obviously)ipi after _every logarithm_
If you want to find just one solution on the other hand, you don't need any additional +2nipi
I agree with you.
C...even with -inf, you can't go lower than e with that
For the other 2, you'll get 0 and 1 as the smallest result, and 2 is above both
2*pi*sqrt(-1) is congruent to 0 mod 2*pi*sqrt(-1)
(Does modulus work that way?)
Yes, this is a sensible way to use modulus.
I think C, probably...it implies e^(e^x)=ln(2) => e^x = ln(ln(2)), now since 0
One calc equation : What's the irl use of finding the volume when we rotate f(x) about a line using the Disc method, the Washer method etc, when we can simply.......
Take the object and submerge it just below a given level of water of volume *v1 cubed units* and see the new reading of reaching the new greater value of *v2 cubed units* , and get the volume of the object as *v2-v1 cubed units* ??? [Done by Archimedes' Principle]
Before watching:
i'll make some tea and then watch
Lol
@@nicebeaver2053 bruh
Please upload a video on "radius of curveature at any point (x,y) on any fuction f(x)..... [Derivation]..
We need power series solution of differential equations MARATHON !!!!😊😊😊
Can you please find the indefinite integral of e^-u^t, where t is a constant?
If indefinite is impossible, then what is the integral from 0 to ∞ of e^-u^t
hey! could you try any eq with this form: a^x+bx^n+c=0
If we search for ALL complex solutions of e^e^e^x = 2, shouldn't we already assume that e^e^x may already be complex?
And if we only search for only SOME solution(s), we don't need the "2*n" but we can simply chose n=0.
Will that result in a different answer, or just add more multiples of 2πn𝑖 that can be consolidated at the end?
Can you do integrals of all trig functions pls
I think the explanation of having to add +2nipi doesn't get at the mathematics behind it. When you apply the complex "logarithm" on both sides the implication from the latter to the former statement will always be true. And indees 2ipi=0 does imply e^i2pi=1. Similarly when finding the solution to an equation taking an arbitrary log will yield *a* solution, just not the solution.
If you add the +2nipi, you can look at it like this in general:
e^x = e^y ==> There exists n in N s.t. x +2nipi = y
This works for the e^i2pi=1 as well here it implies there exists n in N s.t. 2ipi + 2nipi = 0 and indeed, it does with n=-1
Hey @blackpenredpen, I just noticed that the links to your pdfs (eg, 100 integrals) is dead. Is there still a way to access them? Thanks in advance
My first thought is if the answer is A or B, then C also qualifies, dinde e raised to the power of a number not in R will also not be in R.
Therefore the only answers if the question is honest is either C or D. On a standardized test, this is already enough that a random answer is better than no answer.
Then e is around 2.7, so A will have a real solution for x that is less than 1 but greater than 0 because e>2.
To get a value less than 1 of e^x to get a root for B, we need x to have a value less than 0.
To get a value less than 0 for e^x to get a root for C, we need x to be a complex number. That is not in R, so C is the answer.
We have found ln() three times for the equation exp(exp(exp(x))) = 2 so should there be +2*pi*m and +2*pi*k too?
youre my inspiration
I would like to ask if it's possible to solve the following set of non-standard simultaneous equations in x, y, and z for some nice values of a, b, and c (or maybe just a generalised solution)...
x+y+z = a
xyz = b
x^(y^z) = c
Hey! If @blackpenredpen sees this, I have always wondered, if you took the Monty hall problem but made both final doors have an active participant in front of each, making their own decision, do the odds change of winning for anybody? If not or if so, what justifies the changes or lack of change? It bugs me thinking about it.
The setup fundamentally does not work with two active players and three doors in total.
Player A picks door 1
Player B picks door 2
Monty now deliberately opens a door that is unpicked and doesn't have the car. But he can only do that in case one of them has the car. So either he can't open a door *or* he can, but then no one can change.
If you allow for, say, 100 doors instead.
Player A picks door 1.
Player B picks door 2.
Monty now deliberately opens 97 unpicked doors that don't have the car. Now players A and B would both like to switch to the unpicked, unopened door. The car is with p=0.01 behind door 1, with p=0.01 behind door 2 and with p=0.98 behind the unpicked door.
Before watching:
eeeeeeeeeeeeeeeeeee
Hello, I'm trying to calculate the quantity
\int_{0}^{\frac{\pi}{2}} \exp(-y\theta) \cos^{x-1}(\theta) \, d\theta which is a kind of generalized Wallis integral, if anyone has an idea.
Is there a general form for the derivetive of x^^n, like for x^n there is the powerrule?
Very cool video is all I’m saying
Can you do a video on how to find the inverse of (1+1/x)^x=y? I am really curios because i know it exists but I don't know how to get it even with using the Lambert function
Is there a video clip for new viewers with an explanation of what the pen switching is about and why that's his style?
yeah
Check the name of the channel, lol.
All my college profs did that. It's nothing unusual.
Before watching ...
I saw the answer in the comments before watching the video lol
Hello bprp! Can you solve this problem from russian SDA (School of Data Analysis)?
Graph F(x)=lim as n goes to inf(nth root(1+x^n+(x/2)^n)
it seems overcomplicated (btw WA cant solve it), but turns out its not, haha
So blackpenredpen and a bluepen
Sir! I have a question for you (pls make a video on it if u like)
we know that 1^x=1 for all real x; but I have accidentally proved x to be a rational and not real. pls find the fault in my proof. Here goes:
first lest take a general case 1^x=a
therefore: x=ln(a)/ln(1) = ln(a)/ln(e^i2nπ) = ln(a)/i2nπ = -i ln(a)/2nπ {n belongs to Z}
now for 1^x=1 we sub in '1' for 'a' and we get : x = -i ln(1)/2nπ = -i (i2kπ)/2nπ = 2kπ/2nπ = k/n {k and n belong to Z}
therefore x=k/n which is the definition of a rational number.
Pls find a way to fix this!! It will be appreciated.🙏
The very first 'therefore' is incorrect. You are dividing by 0 (because ln(1) = 0).
@@thetaomegatheta it’s not only zero
Zero is just one of the solutions
e^2npi is general solution
According to you which is the hardest question u have ever solved.. make a video on it ❤
C. The natural log of ln(2) is negative, so its natural log is not a real number.
It's one of those times where it's important to actually compute the answers, rather than just finding an exact solution.
One might also say that this is "one of those times where it's important to actually compute the answer" before posting a comment! 😛
Because if you calculate ln(2) it turns out that this is *not* negative after all.
@@jensraab2902 ln(2) is positive, but _the natural log _*_of_*_ ln(2)_ is negative
@@jensraab2902 you may say this is one of those times where it's important to actually read the comment before responding to the comment - he says that the natural log of ln(2) is negative - that's ln(ln(2))
@@XCC23 Good point! 😁
How to integrate x^2a^x by di method,pls solve this question
Third one can't have a solution befause
sol. for A is ln(2) which is positive but less than e
the natural log of something less than e is negative → ln(ln(2)) exists but is negative
ln(ln(ln(2))) is therefore undefined
C does have solutions, but they are all complex.
Funny the ln(ln(2)) is really close to (sqrt(3)-1)/(-2).
Didn’t use ln at all.
As e^0 = 1 < e = e^1, the exponent is between 0 and 1. Which is possible for A and B but not for C as e^e^x > e^0 > 1
Nice!!
Nice!
Isn’t the value dependent on what n is? if n can be any integer does that mean it has multiple solutions?
What 'n' are you even talking about?
EDIT: if we are talking about the solutions for e^e^e^x = 2, then yes, n can be any integer, and yes, there are infinitely many complex solutions there, none of which are real.
IMNENSE THANKS PROF
Couldn’t we say all the fuction are bijection and all have a limit lower than 2 in -infinity and all go to infinity in + infinity? So D would be the answer?
My bad e>2😂
graph y = e^e^e^x and it becomes obvious
I assume that C should be the answer
This time I had to watch the video.
The whole situation is complex, IMHO the solutions are as much real as they are imaginary.
Why don't you have to include "+2 n pi i" when you take the logarithm of a positive number, too?
Your channel is so good I actually had to UNSUBSCRIBE since I keep getting distracted with how interesting these are and will spend hours losing myself working out the solutions.
Oh, I missed the “real” in the title.
I was thinking, “well, the image of exp (as a function from C to C) is everything other than zero, so for all but finitely many values of y, there will exist a z such that e^e^e^…e^z = y , right?”
This reminds me of an old joke told by one of my professors. e^z assumes all values except 0. Thus e^e^z assumes all values except 0 and 1. But it is an entire function, so the Picard's theorem implies that e^e^z is constant. ;P
@@adayah2933 Is the hole that e^e^z actually does take on the value 1, because 2pi i (for example) is in the image of e^z ?
(Uh.. specifically at, z= ln(2pi) + i 2pi/4 )
Nice joke
@@drdca8263 Yeah, that's correct.
Bro what is your educational qualifications?
can you find all roots of "x^8-8x+1"
yeah no roots .
as there is no equation .
Can you solve x! = 3?
First i thought ln2>1 for some reason
Shiiiiish i missed it ......
4:33 💀
Confidence>Arnold on stage
U look cooler than Sam Sulek with a pump😂
That was phenomenal.
U sound like Jensen Huang😁
ln(ln(ln(2))) = ln(−ln(ln(2))) + πi, which is not a real solution.
i wanted to say d but c felt like a trap
c
Before watching
Option c tends to e for x = -infty and must have a positive derivative hence it can never be 2 (with x real ofc)
You forgot something. It's actually:
e^e^e^x=2
e^e^x=ln2+(2n+1)πi
e^x=ln(ln2+(2n+1)πi)+(2n+1)πi
x=ln(ln(ln2+(2n+1)πi)+(2n+1)πi)+(2n+1)πi
I'm too lazy to calculate the complex lns but you get the idea.
Your on the right track, but the 2nd line should be
e^e^x=ln 2 + 2nπi, n∈ℤ.
The (2n+1)πi only appears when you take the ln of a negative real number.
@@MichaelRothwell1 Ah yes
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I have been watching you for a very long timw
helo
ln(x)
ln 2 = 0.693
ln(ln 2) = -0.366
ln(-0.366) = -1.004 + 3.14i (not real)
Do not use equal signs as they are not equal.
@@robertveith6383 r/physics
I wish I was on instagram lol
ln(2) ln(ln(2)) ln(ln(ln(2))) does not have a real solution
Your third line does not make sense, because it is neither an equation nor an equality, so it cannot have a "solution." It is a different question whether it has or not has a real value.
Easiest question
No, it is *not* the easiest question.
Jesus loves you soooo much ❤️
Jesus does not know who I even am, nor who you are. Why would he?
And does he love criminals as well?
Fuck jesus
Cool, I'll add him to my fan club! can he do my dishes?
@@satyam9267 do not be disrespectful
@@OmnipresentPotatowhy not??
2 < e ⇒ ln(2) < 1
ln(2) < 1 ⇒ ln(ln(2)) < 0
ln(ln(2)) < 0 ⇒ ln(ln(ln(2))) ∉ ℝ