Solutions to x^y=y^x
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- čas přidán 21. 08. 2024
- We will solve one of the most interesting and classic exponential equations x^y=y^x
We will use a parametrization to find all the solutions to x^y=y^x
Here's how to solve x^2=2^x (ft. Lambert W function) 👉 • ALL solutions to x^2=2^x
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Find all solutions to x^2=2^x (ft Lambert W function) 👉 czcams.com/video/ndA0sF_0Rwk/video.html
X € R /X#0
Your ability to effortlessly switch between markers is majestic.
Thank you!!!
Nice
Manny Paul A skill you can’t help but learn when you teach lol
@@DarthAlphaTheGreat
Xxx
@@amirnuriev9092 was
It's interesting, that if You draw a graph showing x^y=y^x and allow x=y, You get some curve and a ray y=x and the ray intersects the curve in point (e,e). :) Awesome!
Yay!!Yay!!
thus it shows every possible x and y value where x^y = y^x
I think the more interesting potion of the graph is the discontinuous section where one of X or Y is negative.
OMG this is why I love the internet
It should intersect e,e and pi pi too
It should intersect every point on x=y
I love these no-effort thumbnails, please keep them going 😂
Casey Roberson lol : )
They legendary.
@@15schaa They're*
@@RubyPiec shut
@@wilfriedsteinbach8700 up
Use this to impress girls. LOL. I’ll let you know if it works for me.
GreenMeansGO : )
LOL
How did it go?
It works with girls!
@@davidbrisbane7206 what age-group ?
If you let t = (D+1)/D, where D is integer, you can find all solutions that do not contain radicals. Example: t = 3/2 gives x = 9/4 and y = 27/8.
This yields the sequence of ordered pairs {(2, 4), (9/4, 27/8), (64/27, 256/81), (625/256, 3125/1024), (7776/3125, 46656/15625), ...}, which lists every rational solution. A simple way to generate it is:
x = (1+1/D)^D
y = (1+1/D)^(D+1).
Thus evidently the rational solutions, when ordered in this way (or equivalently, by the size of the denominator when expressed in least terms) approach the irrational solution x = y = e, where the two parts of the curve x^y=y^x intersect.
Very nice! This certainly gives rational solutions, since 1/(t-1)=D is an integer. Probably it gives all rational solutions, but this requires proving.
Parametric generator for x^y = y^x, wow
wow
Wow
I was like wow when I saw it too!
I took 10 courses in math in college en route to a math apps minor and your videos still amaze me.
Here is what I did. First, I took the log of both sides. Thus, yln(x) = xln(y). Then I moved both sides over such that ln(y)/ln(x) = y/x. The problem now looks very simple, its simply saying that both side should be of the same ratio. I then assumed some parameter "c" between both sides of the equality. (We possibly could have done this at the beginning, but that way would have been very difficult). So now we have ln(y)/ln(x) = c and y/x = c. Then we use the linearity of these equations to make our lives very easy. ln(y) = cln(x) and y = cx. Now we simplify the first equation, y = x^c. These two equations imply that x^c = cx. Ah, now both sides have the same base, we can solve for c. x^c * x^-1 = c simplifies to x^(c-1) = c. Taking the (c-1) root of both sides simplifies the equations to x = c^(1/(c-1)). And there you have it! Plug a value into c, and it will yield the x value you need, then multiply that x value by c to get the corresponding y value. This forms the solution space of y^x = x^y
"Don't be too crazy.." *maniacal cackle* oh the numbers I'll produce!!
The most fun part is when you plug in all the complex numbers and see that they all work. I plugged in t = i and not only did it work, but it also generated a real number. Fabulous.
"Don't be too crazy"
Me: puts t=1 *cackles maniacally*
Actually y^x is always equal to x^y if you can write y^x or x^y as the (b-1)th-root of b ^ b * (b-1)th-root of b.
If we say b=3, we have 2nd-root of 3 ^ 3* 2nd-root of 3, which is equal to 2nd-root of 27.
And if b=4, we get that the 3rd-root of 4 ^ 4*3rd-root of 4 = 4*3rd-root of 4 ^ 3rd-root of 4
I actually don’t understand how it takes 13 minutes to proof that x^y = y^x if x=squareroot of 3 and y= squareroot of 27
@@daniledenial you just wrote the same thing that in the video. And the goal wasn't to proof x^y = y^x if x = sqrt(3) and y = sqrt(27), but to solve the equation for real x and y with the condition that x is not equal to y.
@@Amoeby Yeah, but there Are More real Solutions
@@daniledenial of course. But that was the example in the video.
Are powers commutative? Lets just try some values :-P
1, 1 --> 1^1 = 1^1 = 1 --> yes
2, 2 --> 2^2 = 2^2 = 4 --> yes
2, 4 --> 2^4 = 2*2*2*2 = 4^2 = 4*4 = 16 --> yes
2, 3 --> near enough
maybe something a bit less round
sqrt(3), sqrt(27) --> yep
Powers are commutative :-P :) Indisputable proof.
Alan Tennant a simpleton's fallacy.
"indisputable" lol
Alan Tennant love it!!!!
a^(ln(b))=b^(ln(a)) , this way you can commute ...
That's just cherrypicking XD
I just let y be dependent on x, so that y = x^p. We get x^(x^p)=(x^p)^x.
x^(x^p)=x^(px).
x^p=px
x^(p-1)=p
x=p^(1/(p-1))
Plugging in any value for p we get solutions
Adrien Grenier you get the same thing
EDIT: Ohhh, okay I completely misunderstood what you were trying to do.
He knows that. He is looking for values where exponentiation is commutative, that's why he assumes that x^y = y^x (associativity is something else btw)
But how is x^(x^p)=x^(px) equal to x^p=px?
@@nikogruben9573 Both side taking logarithm x-based
This was very high on the "cool thermometer"!
In college in the late 1960's, a few of us math majors were investigating solutions to that equation, just out of curiosity.
While we concluded that (2,4) and its symmetric partner (4,2) give the only integer solution where x≠y, we never hit on this parametric formula.
We did find that if you graph the solution set in the first quadrant, it consists of the line y=x (obviously), along with a curve that loosely resembles the rectangular hyperbola xy = 1, but shifted by (+1,+1); and that the two lines intersect at (e,e) [which is also pretty cool].
Of course, the graph is symmetric about y=x; and as either x or y → ∞, the other → 1⁺.
So this means that the curve mentioned above, more closely resembles
(x-1)(y-1) = (e-1)² ≈ 3
Fred
ffggddss thanks Mr. Fred! Part 2 is coming soon : )
You're welcome! I'll be on the lookout for that. It *is* kind of an intriguing topic.
And having put it aside so many years ago, I was glad to see your parametric solution which, for one thing, makes it a whole lot easier to plot...
Fred
To me, a highschool student who has only done a watered-down version of Calc 1 in a course, because my country doesn't care about Math, I saw this problem as impossible from the title. But as you started with the second solution, I realized how elegant and connected Math can be sometimes.
It just shows how well you can actually communicate your thoughts and explain them well. I'm more of a physics guy, but a deep connection to Math is quite important to me, not just getting something right. You've helped me through that. Thank you for your efforts, and hopefully, with enough practice as it is, I'll be able to view seemingly impossible problems the same way you do!
Totally possible!
Where do you live?
@@diptoneelde836 he lives his own Dreamland where he thinks his own short comings are his country's mistakes. He needs psychology not maths.
@@pigeonlove you have serious problems
plo Judging by the fact you went out of your way to be a douchebag, I think you need psychology.
Nah nah, I agree, absolutely zero countries can function without maths, maths is a quintessential subject in every country for basic functionality over stock exchanges to expeditions, no country doesn't value maths, either the original post is lying and he is just bad and not passionate about maths, or he lives in the tribelands of somalia
Just when I thought there was nothing left for Asians to beat me at, this dude starts writing with two pens in one hand.
never knew this was an asian thing. when i was in school holding a pencil and pen in one hand was normal.
You can't beat us Asians. No one can.
@@alvachan88 lol never seen anyone do that except this guy
Musicians rule: there is always an asian better, and younger than you. Whether you are asian or not.
7:55 That’s what my old maths teacher in high school used to say, when we would suggest a harder way to solve a problem.
If you introduce y = x^t instead of y = tx @13:58 you will get to the same place just as fast but maybe just a bit easier. Good videos, very enjoyable.
The equation e^x=x^e only have the solution x=e. That's the only positive number with this property.
Yes! : )
personsname0 The equation 71^x=x^71 also has a solution x=1.06609898594648539. My equation doesn't have another solution..
thought you were talking bout integer solutions, serves me right for being a smarmy
personsname0 of course I talked about the integer e ;)
lol... oh man, think i might be actually losing my mind
Really interesting. I really like the explanation. Though, when you raise [x^(tx)]^(1/x) x≠0! A big shout out from Argentina!
Bautista Bauto98 thanks!!!!
Actually, one of the solutions you find in your first method is the value y itself (in your example x = 3 is a solution), which you excluded. On way to study the number of solutions of this equation is to rewrite it by taking logarithms (you get y ln(x) = x ln(y) ) and rearranging to a more symmetrical : ln(x)/x = ln(y)/y. So you have to study when the function f(t)=ln(t)/t takes the same value twice. Since f is increasing on ]0, e] from -∞ to 1/e and decreasing on [e,+∞[ from 1/e to 0, it can be shown that for any x in ]1,e[, there is exactly one y in ]e,+∞[ such that f(x) = f(y).
What a great introduction to solving equations with parametrics!
Best way to impress your girls 13:17
Bryan Hart an ad?
I don't get it
Best would be to clear your acne...
looks fucky and here's sexy
x^y=y^x is one of my favourite equations
Yay!!
I actually thought about this problem once, and I was considering natural number solutions for x and y, and I got stuck. After you showed the function x=t^(1/t-1), I graphed it and found that it asymptotes to x=1 as t->+infinity [(+infinity)^0=1 perhaps?], and to t=0 as t->0+. The graph only crosses one lattice point at (2, 2). x is already smaller than 2 for t>2, so no more whole number solutions can exist for t>2. And t=1 is undefined, the limit approaches e=2.718... as t->1 t^(1/t-1) = n->0 (1+n)^1/n = k->+infinity (1+1/k)^k. Therefor, 2^4=4^2 is the only natural number solutions for x^y=y^x.
Jack Ren 2 is the only integer solution as t-1 | t can only happen if and only if t = 2
For x=t^(1/t-1), t=2 gives the only whole number solution of x=2^[1/(2-1)]=2^1=2.
For y=t^(t/t-1), t=2 gives the only whole number solution of y=2^[2/(2-1)]=2^2=4.
2^4=4^2 is the only equation with whole numbers in the form of x^y=y^x, x≠y.
The equastion x^y=y^x on his own is pretty easy. The answer is x=y. But with the data he gave it is a realy hard equastion.
Great video! This is one of my favorite equations of all time.
Try to prove that 2^4 = 4^2 is the only Integer solution in the next video. It's a fun proof
Even tho it's fun it has absolutely no meaning from what i can tell... no real problem would lead to this equation, i guess?
y = x^x or x = y^y
It will work
Please write it, it seems interesting - one of those thing's u have a hunch of but just find difficult to prove. He accidentally made a mistake in saying 3 & 27 work. Thx in advance
Actually no. It might seem like it, but actually see it to completion. Example case (3,3³=27): 3²⁷(3 multiplied 27 times) is NOT equal to 27³(three 3s multiplied 3 times, hence 3 multiplied 9 times)
It all comes down to the solutions of that xᵗ = tx, which I believe to have only (x=2, t=2) as the only integer solutions, related to 2² = 2(2). 35cut may know how to prove it
Theo Ajuyah oh my bad thx for correcting me
In another video u said never trust Wolframalpha
0:54 *P O W E R*
me: oddly satisfying video
others: nerd
r/iamverysmart
@@fyukfy2366 who else watches this besides nerds though?
@@fyukfy2366 totally what I was thinking
@@tweedyburd007 this is not middle school anymore
@@tweedyburd007 true..
9:26 At this point, you divide by t before plugging in x, but then you multiply by t before solving for y. Can't you just plug in x directly and skip a couple steps here?
L1N3R1D3R Ahhhhhhhh loll!!! I didn't see that
Hahahaaa
Haha yep!!
Noticed that as well lmaoo
This guys enthusiasm as he takes us through his story brings unexpected amounts of joy into my life
as always, a very satisfying answer
: )
Thumbs up. That y=tx relation was excellent
1:09 the color of the markers on his shirt are the same of the colors of marker he uses.
What’s cool is if you plot x^y=y^x you get what looks like two graphs super imposed on top of each other. You get the line y=x for obvious reasons, but you also get this asymptote like bit. But from that you can show that 2 and 4 are the only integer solutions (I’m only considering real numbers, not sure if you could have some complex number solutions with integer real and imaginary parts)
8:58
I like that trick. Idk why I haven't thought about it. I might use it on a logarithim question
That impressed me a lot! Seeing such equations solved makes me feel kind of satisfied...
I thought about it, when t < 0, it does not work, x^y will be negative and y^x will be positive, but they have the same decimals. You just add |these| and its okay tho, its a piece of art!
btw if you do this with natural logs:
x^t-1 = t
take ln of both sides
ln(x^t-1) = ln(t)
put power to the front
t-1ln(x) = ln(t)
divide both sides by t-1
ln(x) = ln(t)/t-1
put e as a base to both sides
e^(ln(x)) = e^(ln(t)/t-1)
cancel out e and ln
x = 1/t-1
Hi. Thanks for your math videos! I have a question: around 7:25 you make a simplification which basically is: if a^b = c^b, then a = c. But if b is even, a = -c is also a solution, and this simplification may make you lose solutions. So the proof would have to be completed with the possibility that a = -c and b = 2k, with k an integer, right?
How you can divide by x when the x belongs to the |R?
Well, there obviously are no (sane?) solutions for x=0 anyway, so...
You can try to find integral solutions to the equation with some different approach
Assuming that x,y are positive we can write lnx/x = lny/y or f(x)=f(y) (1) , where f(x)=lnx/x. Through the graph we can see that (1) has infinite solutions (x,y) if x belongs to (1,e) and y belongs to (e, inf.) and vice versa.
Haces la matemática más fácil y entretenida... Más entretenida de lo que de por sí, ya es.
Muchas gracias!
setting y = tx to x = y/t was redundant lol
AniPrograms Yeah, That part was totally unneeded. Still a great video.
You have an anime profile picture
@@orcishh setting y = tx to x = y/t was redundant lol
Now its a valid comment thank me later
@@ilprincipe8094 but this kid has an anime profile picture
@@orcishh goddamnit you are right
Using our beloved W function, we can solve for x with given y>0:
x=e^(-W(-ln(y)/y))
Note that -ln(y)/y is between -1/e and 0, which means the W "function" has actually two different values, one of which gives us the boring x=y result.
You could save a step or two in the third column by starting at y= xt. Also this parametric form shows that the only integer solution occurs at x=t=2 , y=4. Also note that replacing t with 1/t swaps x and y. So all solutions can be found choosing t>1 ie y>x .
Further insights may be obtained by considering the graph of ln(t) against lnx and lny
I feel so smart that I understood what you did in this video
: )
I liked this one a lot!
The solution to this problem for natural numbers goes like this
x | y:
1^(1^1)
2^(2^2)
3^(3^3)
4^(4^4)
5^(5^5)
What a lovely pattern.
Another amazing thing is that if you graph out y^x = x^y with a graphing software, 2.718 appears :0
if you solve this problem without parametrics, but instead through a first derivative function you get : dylny=dx, and as you know value for e is 2.71. I just came across this problem, and in another method, I am encountering a contraction to this solution -- y= x^(t/t-1), therefore x-->t and assumption y=tx may not be correct. (Honestly, makes a whole lot of sense given that rate of change is not linear)
I usually watch some videos of yours, but this one seriously made me think "holy shit this is genius" lol
: )))))
Really cool vid this makes more excited to learn calculus although most of the questions probably will be mostly the daily grind type unlike this one
4:03 That AaaaHaaa caught me off-guard ngl
I have a story with this problem. I had a crush on a girl I've met days ago and I discovered that she had a boyfriend which got me very sad. To distract myself I started thinking about the powers of two and noticed that 2^4 was equal to 4^2 and I was shocked. Decided to try to solve it. Spent two months trying to solve it. Completely forgot about the girl. Since then I have a new crush : solving math problems. Thank you for being an inspiration to me and helping me pursuit my dreams related to math. I'm in high school but I already know calculus because I loved watching your series "math for fun". Love from Brazil
4:09 - i prefer to use ; instead of , between x and y value. For a moment i though that 2nd answer is x= 3.27 and not x=3 y=27
Unlike Europe, Asia and Americas generally use "." For decimal points, like 3"."27 and rarely use "," (except for grouping large numbers like 2,000,000) so yeah it's a bit different
its just that x=3 and y=27 dosnt work at all or i dont get it... 19683 =/= 7625597484987 witch whould be the results for x to the power of y and y to the power of x in this case
@@gepard1983 I was confused too, but I figured it out. Those points are the two solutions where x^3=3^x. The 3,27 point is the trivial one, where 3^3 = 3^3=27. The other one is where 2.4781^3 = 3^2.4781 = 15.2171.
I don't know if you have done a video on this, but you should do a proof that sinh(x) and sinh inverse only intersect at the origin (without graphing tools). It was a problem from my calculus 3 class and I was the only one who got it :P
Essentials Of Math no. I don't have it yet. And I think I will need to think hard on that.
blackpenredpen thinking hard is the best! I'll look forward to it!
Essentials Of Math
I think I will consider f(x)=sinh(x)-arsinh(x) then use IVT for showing a root then use derivative for showing it's always increasing.
blackpenredpen you're on the right track, but there is a much simpler function you can use! :)
blackpenredpen
Think about the derivative of the separate functions sinh and arsinh
There is also a non-parametric solution to this equation. If you want y as a function of x, then x^y = y^x implies y·ln(x) = x·ln(y), which implies ln(x)/x = ln(y)/y. Multiply by -1 to get -ln(x)/x = -ln(y)/y, and this is the form we want the equation in to solve for y. -ln(y) = ln(1/y), hence -ln(y)/y = ln(1/y)/y = (1/y)·ln(1/y), and 1/y = e^ln(1/y), hence (1/y)·ln(1/y) = ln(1/y)·e^ln(1/y). Therefore, we can say ln(1/x)·e^ln(1/x) = ln(1/y)·e^ln(1/y). Now, use the Lambert W function, on both sides of the equation. Then W[ln(1/x)·e^ln(1/x)] = ln(1/y). Keep in mind that W[ln(1/x)·e^ln(1/x)] = ln(1/x) when you take the principal branch, but alternative branches give other solutions, hence why we keep in that open form. That is the key to acknowkedging the fact that solutions other than x = y also exist. y = e^(-W[ln(1/x)·e^ln(1/x)]).
7:26 This is a big mistake. Not only you assumed x isn't 0; but there's also many exceptions , such like (x^2)^(1/2)=((-x)^2)^(1/2). By your simplification, x=-x. See where you are wrong?
👏 this is so good; so good
#U deserve an applause
Charles David thank you!!!!
The 1's cancel out each other. -1+1=0
You can also write x as a^b and, then you have x^y = (a^b)^y = a^(by) and it's your another form of this, e.g. sqrt(27)^sqrt(3) = (sqrt(3)^3)^sqrt(3) = sqrt(3)^3sqrt(3) = sqrt(3)^sqrt(27).
Special and unique cases:
2^4 = 4^2
2^3 +1 = 3^2
"You can use whatever t you want" mmmh let's take t=1
Did you forget that y ≠ x?
Максим Быков yeah you’re right. If t=1 then y would equal x which isn’t allowed
@@disc_00 check it out again
If t=1 ,then it would be a 1/0 as x's exponent
Think first,comment after
@@tylerchristensen7480 @Максим Быков check it out again
If t=1 ,then it would be a 1/0 as x's exponent
Think first,comment after
@@mohammadfahrurrozy8082 Yes by the limit as t tends to 1 is still defined, this gives x = y = e.
There is also an explicit formula whit which you can find the y for a given x:
If y^x = x^y, then y = -x*W(ln(x)/x)/ln(x).
This can easily be verified by trying to transform the right expression into the left one.
PHScience What is W in your equation?
Jonas the lambert W function
I was thinking of the same. But I thought it wouldn't make sense in this context. thank you
Never logarithm when you can find another way.
What's the W function
13:04 Sets t = i ; sees into the future.
Carefully picked rational values lead to nice solutions. Say, anything of the form t = (a + 1) / a. Take a = 3. Then we have x = t^(1/(t - 1)) = ((a + 1) / a)^(1/(1/a)) = (a + 1)^a / a^a, and similarly y = (a + 1)^(a + 1) / a^(a + 1). So they look kind of messy, but plugging in numbers makes them much simpler. (2, 4), (9/4, 27/8), (64/27, 256/81), and so on. You could play around with other parametrized rational solutions that happen to work well and see what you get.
9:27 You divide by t and then multiply by t one step later ;)
y was already isolated :) Still, cool video!
i have a question sir...in this case we assumed Y=tX, so there can be more linear equation to be assumed, and that's how we can get different expressions for the variables.don't we?
Everyone needs to learn this formula. Its great
11:53 using a number “b” for t OR the reciprocal of that same number (1/b) will generate the same x and y values. take t=2 for example it will generate x=2 and y=4, now use the reciprocal t=1/2 and you will have x=4 and y=2
y=x^t and x=y^-t
what about x^n+y^n = x^y for what x,y,n is it true for positive integers? what about negative ones? btw: can you use a head microphone so you do not have to hold 2 pencils in one of your hands?
Me in precalc cp just learning about trig functions binging your videos: Wow, I wish I knew what was going on.
Increasing exponent equals decreasing exponent real value 3. Three tangents to a curve meeting point.
f(x, y) = f(y, x) on Desmos usually gives an identity line.
I've recently had to solve sin(x)^cos(x)=cos(x)^sin(x) on my math exam 😀
Hmmm, x = pi/4 +2npi?
@@blackpenredpen yep :D
Your accent has improved a lot .
Keep up the good work.
Btw, you solved a parametric system of 2 equations.
Remark
At one step, you raised an equality to power .. but you loose the equivalence.
Ex:
A^2= B^2 implies A=B or A=-B.
If you take any number for y, you can always take x=y^y!
because you will have y^y^y on bouth sides...
like 3 and 27 as shown in the graf-solution, you can do it with 5 and 3125 or any number
Why not take the natural log of both sides and solve implicitly?
Micah Beiser it wont do anything
Take the natural log of one side over the natural log of x or y and solve from
log(x^y)/log(y)=x
BLACKPENREDPEN #YAY
WE LOVE YOU
Thank you!!!
We cannot cancel ^x(1/x) as it is not valid if x is even, like (-2)^2=(2)^2 and here we cannot cancel the power directly. So we should only get solutions for odd x.
In order for x^y to be well-defined for arbitrary real y, we must have x>0. So, he should have made that assumption at the beginning: x>0, y>0.
If you set x1. Setting t>1, you can prove there are no integer solutions by showing t^(1/(t-1)) only has an integer solution for t=2, which is EZPZ. Pretty cool : D
Yes we want a video about pytagorian triples #yay
Yahya Hamid I second this
Yep
XaXuser it is easy man
Let k and n be any positive integer
A = k^2 + n^2
B = 2kn
C = k^2 - n^2
XaXuser what to proof here
Νικος Μανε what !? If we work in the set of positifs integers the your A will be greater than your C while the C is the hypotenuse?
Bro’ don’t forgot X & t are different to 0 (cause you divide by X & t)
this is obvious in case of x because if we allow it to be 0, then 0^y = y^0 => 0 = 1, a contradiction
you don't need to know 2^4 = 4^2
Just take log on both sides, and re-arrange a bit, you will get x/log(x) = y/log(y)
Notice that LHS and RHS are symmetric. So you only need to consider one side.
Notice that again, on either side, it is a ratio.
That means you can multiply any non-zero constant to both numerator and denominator, and still, the number is unchanged.
Say, for a non-zero constant k, you have x/log(x) = (k*x) / (k * log(x)) = kx / log(x^k)
which means for any x and constant k, if you can have kx = x^k, then {x, y=kx} will be one of the solution.
I can't thank you enough! You're an excellent teacher.
9:53 Why do you divide by t and then, after substituding, multiply both sides by t??😂😂
SOOOOOOOOOO COOOOOOOOOOOOOL YEEEEEY! Thanks for being awesome with math
this is my favorite channel. I already loved math but on this channel, I learn things I love.
You already had an equation for Y with Y isolated. Why bother dividing Y by T if you were just going to multiply T to both sides anyways?
Aside from that, thanks for the video. I tried to do the ln and got stuck. I thought I could do lambert W function but couldnt find a way to get it into the form once I had x^(1/x) = (#)^(1/#)
This guy is a sorcerer! A sorcerer I tell you!
is this considered pre-calculus? how important would parametrics be going into calc 2?
I think it's do-able in precalc. Parametric equations are useful in calc2, 3, and more : )
thanks! and are you at UCB? do you teach there? would be fun to have you
Note: you can divide by x because x≠0 otherwise y=x×t=0t=0=x with is in contradiction of x≠y
The complex numbers makes everything more fun
I hve learn new thing today, thanks
Great solution!
(Should I do this? Eh, it’s a rare opportunity. No offense though, you’re still WAY smarter than me. See below for his mistake)
At 9:30 just multiply tx through; that’s exactly what you did after wasting 30 seconds and adding unnecessary confusion.
To be honest, I am not that smart.
What's interesting is that if x^y=y^x, then x and y are proportional to their logarithms. In fact, we have
x^y=y^x
y ln x=x ln y
(y ln x)/(ln x ln y)=(x ln y)/(ln x ln y)
x/ln x=y/ln y
Your second solution is nice for proving that 1^infinity is indeterminant. As t gets very large (approaches infinity), 1/(t-1) will approach 0 => x will approach 1, and t/(t-1) will will approach 1, which means y will approach infinity. You'll get a conclusion that 1^infinity = infinity, which is an alternative to the intuitive solution that 1^a = 1 for all a.
really nice video! but can you also solve x^2=e^x with it?
Philipp Huber parametric equations is used to solve implicit equations
Philipp Huber while you can still solve it using the lambert w function
x^2=e^x
x^2 e^(-x)=1
xe^(-1/2x)=±1
-1/2xe^(-1/2x)=±1/2
-1/2x=W(±1/2）
x=-2W(±1/2)
while W(-1/2) is complex
so the real solution of x is-2W(1/2)
Philipp Huber en.wikipedia.org/wiki/Lambert_W_function
無謂 I generalize it with this formula
x^p=b^x
x=-pW(-ln(b)/p)/ln(b)
無謂 ok this is a little too high for me
Don't worry, be commutative! #yay
If you say (3sqrt(3))^sqrt(3) = sqrt(3)^(3sqrt(3)), you can say the same thing, where you only use the number 3 for t=3.
Are there other solutions ?
a^x=b^x=> a=b ?
Wrong if x=2 and a= -b0