Solving popular exponential equations

How to the exponential equations x^x^3=3 and x^x^3=36.
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#math #algebra #mathbasics

[@cosmicvoidtree]

To solve x^x^3, it’s rather simple. Cube both sides to get (x^3)^(x^3). Use the x^x solution plug in x^3, and take the cube root to cancel the cubing. The end result is e^1/3*W(3ln(x))

[@redroach401]

this method should only be useed for the 2nd question because for the first question you can just simplify using a^a = b^b to get x^3 = 3 so using cubic identity: (x-3^(1/3))(x^2+3^(1/3)x+3^(2/3))=0. So x=3^(1/3) and whatever that quadratic simplifies to.

[@garyhuntress6871]

Always like things involving exponents. Every time I hear "W", I expect to also hear "fish"

[@nokta9819]

6:32 for this part i used Lambert W function and i saw that we can write infinitely many solutions When x^x^3=a, x= (3 ln(a)/W(3 ln(a)))^(1/3) So then when we look at the part in cube root, we have to get rid of the W function, and we also want that the ln(a) parts to cancel. For this one, i used this rule: W(x ln(x))=ln(x) Lets say a=b^c The W function part will be W( 3c ln(b)), so if we want to use that rule we have to equalize 3c and b. When we get rid of the W function with this rule, the ln(b) parts also cancel out and the last expression will stand like x= (3c)^(1/3). We can simply find x^(x^3) one, that will be (3c)^c = x^(x^3) If we want nice expressions for x^(x^3)=a, a has to be in (3n)ⁿ form. So we can find infinitely many solutions like 3, 36, 729, 20736, 759375, 34012224, 1801088541...

[@Osirion16]

I arrived to the same conclusion but without using the lambert W function, I gotta admit, though, that it is a beautiful way to prove it

[@nokta9819]

​@@Osirion16thank you man, appreciate it But I wonder if you could tell how did you do it It is doable actually but how

[@Osirion16]

@@nokta9819 It's been like a year and I don't exactly remember what my thought process was but BPRP only asked if there were any other "nice" solutions to the problem he proposed, it is assumed that every "nice" solution has to be written in the a^a form, which can then solve the equation from the fact that x^x^3 will be cubed to get (x^3)^(x^3) therefore, we are always gonna have to cube the left side and so we will also have to cube the right side, meaning that a^a has to be in the form a^b where 3b=a when cubed, you're gonna get a^3b=a^a but that also tells you that a has to be a multiple of 3, from there, you can find infinitely many solutions from the simple fact that a has to be a multiple of 3 and b has to be equal to a/3 (basically what you said) However, this technique completely disregards other "trivial" solutions to the problem, like, for example, the integer "2" : If x^x^3=256 then you can rewrite the equation as x^x^3=2^8=2^2^3 which implies that 2^2=x^x, meaning that 2 is a solution and it is also an integer But as I said, this is trivial because any integer will get a solution as long as you plug it into the equation first meaning that if you want 5 to be a solution, just calculate 5^5^3 and then set that equal to x^x^3 and you'll get 5 back

[@AaronWGaming]

Basically what he did with the (X^3)^(X^3) is changed it to u^u =n^n Where n is 6 then the issue is he needs X not U... But U=x^3... Take the Cube Root of both sides...

[@noahali-origamiandmore2050]

Any number a^(a/3) will work with the challenge question. Also, you didn't consider that both equations that you solved have two complex solutions (not counting the ones that you get from exponential functions). You can easily find them by just knowing cube roots of unity and multiplying them by cbrt(3) for the first one an cbrt(6) for the second one.

[@user-xi7qv6ti6v]

That just was an amazing mathematics problem!

[@NatoSkato]

6:32 I just speculated by looking at a pattern. You always get a "nice" solution if you say that x^x^3 is equal to (3n)^n. That means 3^1, 6^2, 9^3, 12^4 15^5... all equal to the cube route of 3n. You can do 27^9 and get cube route of 27, or 3. I'm trying to figure out why this is the case, I just thought it was nice. No maths just pattern recognition. edit: just realised someone did it before me, gonna leave it up anyway.

[@mathprodigygamer4637]

(I don’t know if the other comment explaining this told you… but) The reason the patter is like that is because if you have a cube root for x^x^3 as the answer, then you need a multiple of 3 inside the cube root because the top area of x^x^3(x^[x^3]) will give you x^(the multiple of 3 inside the cube root) with then gives you [(the multiple of 3)^1/3]^(the multiple of 3). Then you can simplify that by multiplying (the multiple of 3) by 1/3 to get a whole number, so the simplified version would be: (the multiple of 3)^(the whole number you got). Sorry if this is a bit confusing, but it’s kinda hard to explain it without showing it.

[@SidneiMV]

x^x³ = 3 x³^x³ = 3³ x³ = 3 => *x = 3^(1/3)* x^x³ = 36 x³^x³ = 36³ = 6⁶ x³ = 6 => *x = 6^(1/3)*

[@to2podemosaprender630]

Me gusto mucho... que bacan es el algebra...

[@rki7068]

Where is a good aplce to find these kind of practice problems?

[@aliali-i2z5q]

Wow 😳 thx

[@brawlingpanda4019]

3:52 My caption show me that he said the lamborghini function

[@yairkaz]

for x^x^x...^n = n if n is even it's +-sqrt(n). for example x^x^2 = 2 it's +-sqrt(2). Also those are only the real solutions

[@oenrn]

x^x^x... gets weird for negative values.

[@yairkaz]

@@oenrn gets weird doesn't mean undefined

[@oenrn]

@@yairkaz yeah you're right, I was thinking of the infinite series x^x^x..., which CAN be undefined for certain values. The finite series doesn't have that issue, and if it ends on an even number will always make every exponent in the chain a positive number.

[@yairkaz]

@@oenrn the infinite series is limited with it's values ranging from I think 1/e to e

[@miki3386]

Please Prove the formula X=n^1/n for the very first question🙏

[@oenrn]

x = n^(1/n) Raise both sides to n power: x^n = n Therefore: x^x^n = x^n = n x^x^x^n = x^x^n = x^n = n And so on. Regardless of how many x it will always hold.

[@Zero-xb6ls]

I think you should do lessons explaining parts of trigonometry like logarithm or sine functions and how to use them

[@pneujai]

logarithm is not trigonometry

[@Zero-xb6ls]

@@pneujai bro I don’t know I ain’t learnt it

[@economicsapplication6083]

Answer to your question :- 81 to the power 27. Then x= 3*cube root 3.

[@vfx7t]

Ok thanks, what about Ln ?

[@WerewolfLord]

Well, for x^x^3 = ?, 256 gives a "nice" expression, assuming, of course, that you consider "2" to be a "nice" expression.

[@p12psicop]

729?

[@Oskar5707]

can we find x to the x to the 3 is 256?

[@quasarsareinsane]

yes we can. the answer is 2. x^x^3 = 256 x^x^3 = 4^4 x^2^3 = 4^8 (multiplying both the exponents with 2) x^2 = 4 & x^3 = 8 hence the value of x is 2

[@candlecrown8717]

How many solutions are there for x power x problems?

[@TheBatugan77]

X.

[@brahdavv8371]

This first one is lower attitide

[@Quaizyer]

x^x^3=64 Has a nice solution

[@jpjaye]

I tried solving it and I ended up with x=x that was fail

[@jaykay3624]

x= cubed root (6) appears to solve the equation.

[@mhmaya1439]

when x^x=n^n than provr that x=n

[@TheBatugan77]

XX = Dos Equis

[@casadelosperrosstudio200]

Hmmmm... so x^x^3^3 isn't x^x^9 ? 😮

[@plislegalineu3005]

x^x^3 = 729

[@diegolau245]

Yo read my mind

[@Almashina]

so what is "nice" *x* here? Obviously not *3* because *3^27 >> 729*

[@plislegalineu3005]

@@Almashina fuq I probably did a silly mistake

[@plislegalineu3005]

ah no it's ³√9

[@Almashina]

@@plislegalineu3005 yep, it is really nice root, much better than in video

[@yugam6578]

Nice problems

[@user-zo8sq9pg1m]

おもろい