So many people actually dont know how to deal with the expressions with the lambert w function; The expressions in lambert W's with ln's are actually way easy to simplify; So the question is to solve x^(x^3)=2^(2^(14)) First we ln both sides x³ ln(x) = 2¹⁴ ln(2) And we multiply both sides by 3 to get the same exponent x³ ln(x³) = 2¹⁴ 3 ln(2) And we take the W of both sides using the rule W(x ln(x))=ln(x) and we get 3 ln(x) = W(2¹⁴ 3 ln(2)) And we divide both sides by 3 and exponantiate both sides; the result becomes like follows: x = e^(W(2¹⁴ 3 ln(2))/3) So far so good but everyone used wolframalpha as i see so i want to help with simplifying that So I made a formula with some tricks but it is a bit long so i will just show how to do that Given the expression W(a^m n ln(a)), we move like follows; 1) Set a and n as relatively primes (for an example if n is 6 and a is 2, just break 6 as 2*3 and increase m as exponent) We can see that a and n are relatively primes (2 and 3) 2) We set x as a variable and solve for the equation n*a^x + x = m We can use trial and error method, but if we go more theoretically the approximate solution is (ln(m)-ln(n))/ln(a) If we know that this expression in the W is simplifiable, x will be an integer. So we use the formula for the approximate value; (ln(14)-ln(3))/ln(2); which is approximately 2.222... so our integer is 2. We solved for x, what is the next step? If the given expression W(a^m n ln(a)) is simplifiable, we set a variable x and solve x for the equation n*a^x + x =m The simplified version of the expression looks like (m-x) ln(a) So we know a, m and x Lets substitute; a=2 m=14 x=2 We get (14-2) ln(2) which is 12 ln(2) Lets move on; We substitute the expression as 12 ln(2); e^(12 ln(2)/3), 12 and 3 cancel out and become 4, e and ln(2) cancel out and become 2, our last expression stands like 2⁴ which is 16.
@@CriticSimon It was a very foolish thing to do. He didn't say that he was using the wrong expression IMMEDIATELY as he showed it. He showed it at 0:40, then yammered on and on and on until 2:10 (i.e. 90 seconds LATER), just after he seemed to have finished the WA stuff, but even then he only mentioned that he had used the wrong expression as an aside. He gave no explanation as to what possessed him to use an incorrect expression, other than if you do, then you will get an incorrect result - WHO KNEW!?!?!? His decision to do that makes no sense at all, and I now note that it led to many people pointing out the mistake. He has done exactly the same mistake in another video, but didn't realise that he had done it. He could have done a far better job of it if he really wanted to explain the order of operations for power towers.
Nice work Sybermath! For some reason, I like the second method so much :)
Even in second method you don't need w(lambarts) function
You start with Wolfram Alpha showing (x^x)^3 which equals x^(3x), but your problem involves x^(x^3) which is different.
OK, you corrected yourself at 2:20.
Beautiful question...❤️❤️❤️.
So many people actually dont know how to deal with the expressions with the lambert w function;
The expressions in lambert W's with ln's are actually way easy to simplify;
So the question is to solve x^(x^3)=2^(2^(14))
First we ln both sides
x³ ln(x) = 2¹⁴ ln(2)
And we multiply both sides by 3 to get the same exponent
x³ ln(x³) = 2¹⁴ 3 ln(2)
And we take the W of both sides using the rule W(x ln(x))=ln(x) and we get
3 ln(x) = W(2¹⁴ 3 ln(2))
And we divide both sides by 3 and exponantiate both sides; the result becomes like follows:
x = e^(W(2¹⁴ 3 ln(2))/3)
So far so good but everyone used wolframalpha as i see so i want to help with simplifying that
So I made a formula with some tricks but it is a bit long so i will just show how to do that
Given the expression W(a^m n ln(a)), we move like follows;
1) Set a and n as relatively primes (for an example if n is 6 and a is 2, just break 6 as 2*3 and increase m as exponent)
We can see that a and n are relatively primes (2 and 3)
2) We set x as a variable and solve for the equation
n*a^x + x = m
We can use trial and error method, but if we go more theoretically the approximate solution is (ln(m)-ln(n))/ln(a)
If we know that this expression in the W is simplifiable, x will be an integer.
So we use the formula for the approximate value; (ln(14)-ln(3))/ln(2); which is approximately 2.222... so our integer is 2.
We solved for x, what is the next step?
If the given expression W(a^m n ln(a)) is simplifiable, we set a variable x and solve x for the equation n*a^x + x =m
The simplified version of the expression looks like (m-x) ln(a)
So we know a, m and x
Lets substitute;
a=2
m=14
x=2
We get (14-2) ln(2) which is 12 ln(2)
Lets move on;
We substitute the expression as 12 ln(2);
e^(12 ln(2)/3), 12 and 3 cancel out and become 4, e and ln(2) cancel out and become 2, our last expression stands like 2⁴ which is 16.
2 taff 4 me.
May have to watch the video 16 times.
I think you typed the question incorrectly in Wolfram Alfa, it's not (x^x)^3 but x^(x^3) ...
Very nice!
You used (x^x)^3 instead of x(x^3) on Wolframalpha. It's not the first time you made that mistake.
Did you watch the whole thing?
@@CriticSimon I hadn't, but I see that he dealt with it. It's very strange that he knowingly used the wrong expression, yet published the video.
@@Chris-5318 he used the wrong expression and then he explained why he used it
@@CriticSimon It was a very foolish thing to do. He didn't say that he was using the wrong expression IMMEDIATELY as he showed it. He showed it at 0:40, then yammered on and on and on until 2:10 (i.e. 90 seconds LATER), just after he seemed to have finished the WA stuff, but even then he only mentioned that he had used the wrong expression as an aside. He gave no explanation as to what possessed him to use an incorrect expression, other than if you do, then you will get an incorrect result - WHO KNEW!?!?!? His decision to do that makes no sense at all, and I now note that it led to many people pointing out the mistake. He has done exactly the same mistake in another video, but didn't realise that he had done it.
He could have done a far better job of it if he really wanted to explain the order of operations for power towers.
"3 is the exponent of x but, that's not the exponent for x."
3-T scroat
Using ln you are just complicating everything. Let x=2^t; then t2^3t=2^14 Hence t=4;
x=2^4=16
X=25.39841
16
x=e^(W(ln8*2^14)/3)=16