A Homemade Exponential Equation

Показать описание

🤩 Hello everyone, I'm very excited to bring you a new channel (SyberMath Shorts).
Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡

SIMILAR PROBLEM

If you need to post a picture of your solution or idea:
x^x^3=2^2^14
#exponentialequations #exponentials #exponents
via @YouTube @Apple @Desmos @NotabilityApp @googledocs @canva

PLAYLISTS 🎵 :

SyberMath

Рекомендации по теме

Комментарии

Nice work Sybermath! For some reason, I like the second method so much :)

PKMath

Even in second method you don't need w(lambarts) function

vcvartak

You start with Wolfram Alpha showing (x^x)^3 which equals x^(3x), but your problem involves x^(x^3) which is different.

mbmillermo

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.

nokta