A Nice Equation | Algebra Question | You should be able to solve this

[@MichaelRothwell1]

Thank you for this video which is carefully and beautifully presented, although I don't agree with some of the things said. Whilst it is a valid move to rearrange the equation xˣ=x into the form of the more familiar equation aᵇ=1, I think that more important than knowing the solutions to this equation is knowing how those solutions are found, since the same method can be used to solve similar exponential equations. One method is to use logs, and this method can be used to solve the equation directly (without rearrangement). Here is my solution. We are to solve xˣ=x for real x (a reasonable assumption, I think). We consider 3 cases: x>0, x=0 and x0, we have xˣ>0 (as a real power of a positive number is positive), so we can take ln of each side: xˣ=x ⇔ln xˣ=ln x ⇔x ln x=ln x (we can use the rules of logs here as the base x of xˣ is positive) ⇔(x-1)ln x=0 ⇔x-1=0 or ln x=0 ⇔x=1 (in each case) For x=0, we have xˣ=0⁰=1≠x. So x=0 is not a solution. This conclusion is true even if you think that 0⁰ is undefined. For x

[@MichaelRothwell1]

P.S. I mentioned above that we can apply the rules of logs to powers where the base is positive. It's worth noting that we can apply the rules of exponents where the base is positive and also where the power is an integer (this applies whenever the powers are defined, and the base can be any real or complex number or even a matrix etc).

[@mathserreurs2479]

Thank you for subscribing to my chanel. Je viens de faire pareil. Je vois que tu parles français aussi, cela m'aidera à améliorer mon anglais. Bonne journée.

[@tanmaychopra4239]

Fantastic, I learnt a lot!

[@miegorengq]

I want to ask something, why X^x=x^1 Its same with x^x÷x=1 ? Example, 2^4=2^4, Its not same with 2^4:2=4, 2^4=16, 16÷2=8, 8 not same with 4, so why x^x=x^1 Its same with x^x÷x=1 ?, Please aswer me, Im not understand

[@MichaelRothwell1]

@@miegorengq if we divide both sides of an equation by the same non-zero expression, the equation will still hold. If we divide both sides of the equation xˣ=x by x (for x≠0) we get xˣ÷x=x÷x LHS=xˣ÷x=xˣ÷x¹=xˣ⁻¹ and RHS=1 So the equation becomes xˣ⁻¹=1 Taking your example, which is to divide both sides of the equation 2⁴=16 by 2, the correct calculation is 2⁴÷2=16÷2 LHS = 2⁴÷2¹=2⁴⁻¹=2³=8=RHS Note that x^x÷x, as you wrote, means (x^x)÷x and not x^(x÷x). Similarly, 2^4÷2 means (2^4)÷2 and not 2^(4÷2).

[@tildarusso]

I can work out the x=1 by taking logarithm on both sides instantly, with base of x, log_x(x**x)=x, log_x(x)=1, therefore x=1. But since base can not be negative the x=-1 is however lost.

[@Mj382-d73]

yea but if you take this into consideration you can deduce that |x|=1, meaning you didnt loose the -1

[@theexplosive583]

isn't -1^(-1) is also positive 1? -1^(-1) = -1/(-1)^1 = -1/(-1) = 1 or I have messed up something?

[@theexplosive583]

oh sorry, I found my mistake. I calculated like a^(-n) = a/a^(n) instead of using 1/a^n

[@thenoobyblock1208]

@@theexplosive583 pretty sure (-1)^(-1) is just the reciprocal of -1, which is -1

[@krishnakarthik4752]

No, I think -1 is also a valid solution

[@FitSS369]

bases are equal so we can equate the exponents which is x=1

[@novidsonmychanneljustcomme5753]

True, but how do you obtain the second solution x=-1 then?

[@sathwikmadapati618]

@@novidsonmychanneljustcomme5753 x must be of form 2n. SUCH 2 SECOND questions 🙄man for JEE ADVANCED INDIAN ASPIRANTS..😂

[@HaiLe-wy7dd]

1^2=1^4, bases are equal but exponents are not. Your reasoning is wrong.

[@snipyjodgaming2206]

@@novidsonmychanneljustcomme5753 x power - 1 is also giving 1 so - 1 and 1 both are the values for X We just have to use our common sense

[@maf427]

@@HaiLe-wy7dd Sure, but it works with unknown quantities. Let 1^x = 1. 1 can = 1^1 therefore x = 1 but it doesn't mean x = 1 is the only solution. Now 1 can also equal 1^-4 therefore x is also equal to -4. It's conclusive that x = all real numbers.

[@giorgosanas1608]

🪓 divided by 🪓 minus 🪓

[@daniswaraputra299]

The equal is unreachable

[@dominickdayang1838]

Nice 😅

[@Samiul_007]

Lmao

[@abm5195]

Lol😂

[@disgracedmilo]

Hopefully this is a lighthearted joke

[@pranavsahnii]

Both +1 and -1 could be the real values of x

[@ramking7869]

And 0.

[@mrmsaid7617]

@@ramking7869 nope

[@chaoseros1714]

@@ramking7869 remember anything to the power of 0 equals one which means if u substitute x for 0 it’ll equal 1. I think some people think 0 to the power of 0 is 0 tho so idk if ur that person

[@seroujghazarian6343]

@@chaoseros1714 nah, it's defo 1

[@sandhuekam14]

@@seroujghazarian6343 0 to the power 0 is undefined nor 1 nor 0

[@justabunga1]

Since the domain isn't mentioned here, we have to extend some possible values. This means that we have to take the absolute values on both sides of the equation (both the base and the exponent). In the next paragraph, it will explain why the line doesn't cross one of the solutions. Let's the set the equation as |x|^|x|=|x|. Take ln on both sides, which is |x|ln(|x|)=ln(|x|). Subtract ln(|x|) on both sides, which is |x|ln(|x|)-ln(|x|)=0. Factor out ln(|x)|, which means that (|x|-1)ln(|x|)=0, or x=±1. You can go ahead and check the equation to make sure that is the right solution. Originally, there's not supposed to be an absolute value there since the graph changes. The reason why x cannot be -1 is because of the domain if you graph y=x^x and y=x. The graph of y=x^x has the domain where x>0. x=-1 is out of the domain, but putting in the value for x=-1 is the right solution. Because of this and even though you could plug in negative values for x, we want the graph to be continuous. Let's say you put in x=-1/2. There it goes, the value is not real. Therefore, we cannot have negatives values for x. Otherwise, it keeps jumping the values between real and imaginary numbers. x cannot be 0 be cause 0^0 is undefined, but it does have the limit to be 1 as you approach from the right side. The answer comes from the use of l'Hopital's rule since 0^0 is an indeterminate form. An indeterminate form since there are 7 of them means that the answer cannot drawn from the conclusion meaning that you have to do show more work. To get the answer for that limit, y=x^x=e^(xln(x)). Take the limit of the exponent and then raise that by the base of e. xln(x)=ln(x)/(1/x). Take the derivative of the top and derivative of the bottom separately. Note that we are not doing the quotient rule derivative. This means that (1/x)/(-1/x^2)=-x. The limit is therefore e^0=1. The graph of y=x is easy to figure things out since you already know this from algebra. On the other hand, the graph of y=x^x will have to be done using the calculus method to show where the line increase/decreases (or max./min. pt.) using the first derivative and set that equal to 0 and concavity by setting the second derivative equal to 0. The first derivative of y=x^x is y'=x^x(1+ln(x)). Set that equal to 0 has the value of x=1/e. Now we need to see where the max./min. pt. is. Let's try something between 0 and 1/e say like x=1/e^2. We can tell that the value is negative, so the line is decreasing. Try another value that is greater than 1/e say like x=1. The value is positive, so the line is increasing. Let's find the second derivative, which is y"=x^(x-1)(1+x(ln(x))^2). If we try to set the second derivative equal to 0, you won't find any values for x. However, we have to come up with another method from the equation. From there, x is always going to be positive. This means that it's always concave up.

[@samueljehanno]

😮

[@davidemmanuel9418]

I was expecting an answer at the end 🙁

[@kotarotennouji5238]

@@davidemmanuel9418 he basically said x = 1 not -1 and used logarithms and derivatives to prove that statement.

[@xXxkwexXx]

Hahaha well done bro, I stop the video when he not even started with the log in both sides 😂😂😂 newton will go crazy if he saw this 😂😂 equation is for children, give me the real world , give me the function 😂😂

[@mythsealaes2206]

I wonder if there are any complex solutions...

[@epikherolol8189]

Not

[@tengig7534]

it is possible to solve this equation according to the graph. The graph of the function X is a line passing through the coordinates (0; 0) (1; 1). Function graph х^х have coordinates (1; 1) (2; 4) (3; 9) etc. So, these graphs intersect only at one point (1; 1) therefore, this equation has one solution. Х=1

[@timbomb374]

After looking at the thumbnail and pretty much intuitively figuring X=1and probably -1. I did not expect there to be so many equations to go through to prove it.

[@oZqdiac]

Yes you are right x can also be -1 Proof -1^-1 = 1/-1^1 = 1/-1 = -(1/1) = -1 -1 = -1

[@snikdud]

POV: you solved this immediately

[@michellepopkov940]

By inspection x=1. Divide both sides by x. X^(x-1) = 1. So x-1 (the Exponent) should be zero. So x=1.

[@mathserreurs2479]

x^p=1 with p=x-1 { p=0 and x0: x=1 solution, Or x=1 and p in R, Or x=-1 and p even: p=-1-1=-2 is even so x=-1 is solution } For more examples see: kzbin.info/www/bejne/l3illmyFl9afraM

[@davidbrown8763]

Exactly what I thought.

[@oZqdiac]

x is also -1 as well

[@spinothenoooob6050]

it's so simple, x^x=x because the bases are equal the exponents are also equal therefore, x=1

[@kobalt4083]

There is also another solution x = -1.

[@math4u769]

Take ln of both sides. Transpose then factor. Solve for x.

[@ikirigin]

I solved this by taking the log of both sides, so [ x lnx = ln x ]. Then [ ln x (x-1) = 0 ]. Both point to 1 being a root. Where can you get -1 from this method?

[@touhami3472]

See: kzbin.info/www/bejne/l3illmyFl9afraM

[@sathwikmadapati618]

Use |x|. Always write log|x| when using in an equation to equate it to 0....

[@touhami3472]

For x^(x+2)=x^(1-x) ln leads to (2x+1)ln|x|=0 : then 1, -1 and -1/2 are solutions !!! But true solutions are: 1,0, -1/2 !!!

[@SerialBull]

Divide both sides by ln x to get x = 1

[@Dissandou]

You eliminate negative solutions when you do ln(x) instead of ln|x|.

[@zwz.zdenek]

I originally guessed that y=x would intersect x^x in two places, but I see it's tangent at 1. How would one approach a more general case, such as x^x=2x?

[@xdelqyed1090]

Can u not solve this with laws of index by canceling the two bottom x’s

[@Johannes_Seerup]

x^x = x^1 Since the bases are the same, the exponents can be set to equal each other:) x = 1

[@shirobitoo]

I can see that it’s x = 1,-1 without even trying to solve it. Proving it though… that’s a different story. I think using log would be helpful, but at the end of the day you’ll get log base x of x = x lol

[@pranavtiwari_yt]

But base can't be negative so how solution -1 is correct?

[@creaturekaspar]

a negative base is perfectly legal, -1^-1=-1

[@pranavtiwari_yt]

@@creaturekaspar yes numerically it is but in x^x as a function it is not

[@Rkumar123-k8u]

You can also take log both side

[@Rkumar123-k8u]

There is condition(x>0)

[@Frank-kx4hc]

Why lot of comments are censured in your videos?

[@IoDavide1]

I suggest you to abandon the use of obelus, just use the slash. In a video the obelus is very easily confused with a minus sign.

[@kasskmb]

x^x=x x^x=x^1, equal bases, cancel the bases and equal the exponents x=1

[@prosimulate]

1. With logic.

[@novidsonmychanneljustcomme5753]

True, but how do you obtain the second solution x=-1 then?

[@ruchikarfacts7380]

Olympiad questions 2^pq - 2^rs = 32; where p,q,r,s belongs to natural number then Possible values of p + q + r + s = ? Solution:- kzbin.info/www/bejne/mIO7i2Vjn8aCbtU

[@Gideon_Judges6]

Can't we just take log of both sides? |x| = 1, x = {1, -1} ?

[@BasitCeviriler]

Yes we can, ı solved with logarithm

[@OrDoL]

I believe you can't

[@kekw8105]

I believe you can

[@OrDoL]

@@kekw8105 how can you?

[@ruchikarfacts7380]

Olympiad questions 2^pq - 2^rs = 32; where p,q,r,s belongs to natural number then Possible values of p + q + r + s = ? Solution:- kzbin.info/www/bejne/mIO7i2Vjn8aCbtU

[@juandelacruz9027]

This is quite easy. Just compare the left side and the right side. The exponent on the right is obviously 1 since it's not written, so the answer is 1.

[@ronchristiensantos5501]

x^x = x, since the bases are equal, just focus on the exponents x = 1, and you're done

[@chowdavid1874]

Take absolute value on both sides. The only possibility is that |x| = 1, and can be easily verify that x = 1 and -1 are the solutions. Simple question done in 3 seconds.

[@mohandalansari534]

Infinity^infinity=Infinity and x=±1

[@איתיסמואלוב]

And for those here who also want to solve it *correctly*- apply ln for both sides (we will check later for negative x solutions...) And you can now substract ln(x). From both sides. So ln(x)(x-1)=0 Both ln(x) and (x-1) become zero when x=1. Because ln(x) is valid only for positive values of x- we do the same with u=-x By that we will get u=1... x=-1

[@mappingsworld]

we know this is true because u can multiply 1 to the power of 1 to the power of 1 and it can keep going on infinitely, but IT WILL ALWAYS BE ONE

[@N.N-p9g]

ln-logarithm both sides, then we see immediately in our heads(!) that the or-conjunction of ln(x)=0 and x=1 is the only solution of our equation.

[@VectorJW9260]

x=1 is a solution! I found this out looking at the thumbnail. I know that 1 raised to any (real, at least) power is itself, and so I just plugged it in and it worked out.

[@VectorJW9260]

(Though I did lose the x=-1 like the other people who used other methods)

[@kittisakchooklin874]

That makes sense. Was checking in the rational form, and it simplified to -1

[@thetntmaster3127]

Log base x both sides?

[@epikherolol8189]

But is log base x feasible?? Coz we need to find x so how can we take log base x, i mean we can take but it's not feasible. Take either natural log or log base 10 or any base other than x

[@NamMathProf]

simply apply the power rule, if the base are the same, you only equate the index

[@mr.cracker9317]

Yes, it must be 2 line answer, 1 line question and another line x = 1, 😁😁

[@novidsonmychanneljustcomme5753]

True, but how do you obtain the second solution x=-1 then?

[@mr.cracker9317]

@@novidsonmychanneljustcomme5753 x = -1, this answer can also be matched but I didn't see any correct solution of x^x = x which gives x = -1

[@GanonTEK]

@@mr.cracker9317 If you plot the two functions, the line y=x touches it twice, at 1 and -1. So it is a correct solution. Some graphical calculators struggle with x^x at x

[@mistashadow]

That's because most of it isn't real numbers when x < 0 Sure, (-1)^-1 = 1/‐1 = -1 and (‐2)^-2 = 1/((-2)²) = 1/4 = 0.25 But then we get things like (-0.5)^(-0.5) = 1/(sqrt(-0.5) which is not a real number

[@RhinocerosMovie9]

Wow so Wonderful! Highly impressed with the channel. This girl is super genius. Well done. Keep it up! #SUPERMATHS

[@LarryPanozzo]

Smart people: • It feels like it should be 1 and maybe -1 too. • Does “guess and check.” Yep it’s -1 and 1! • Checks that 0 doesn’t work and values higher than 1 and lower than -1 diverge. Boom ‘solved it’ in 10 seconds! 😅

[@sathwikmadapati618]

USE LOG AND PROPERTIES ON BOTH SIDES to get 1,-1 and consider ln|x| . THEN U WILL GET ANSWER IN 2 SECS. Such simple 2 second questions these are for JEE ADVANCED INDIAN ASPIRANTS man😂😂. IF YOU TRULY WANT TO DO TOUGH MATH SUMS THEN TRY SOLVING JEE ADVANCED 2022 MATH PAPER IN 1 HOUR.

[@oZqdiac]

@@sathwikmadapati618 dude chill this method works just fine for this question as well

[@sjejnwbww]

-1 is no solution of this equation

[@oZqdiac]

It is Here’s proof -1^-1 can be written as 1/(-1^1) -1^1 = -1 so this is now 1/(-1) 1/-1 can be written as -(1/1) 1/1 is 1 so this is now -1 And -1 = -1 So there

[@SousouCell]

X^x is only définid in ] 0 , + inf [ Therefor the solution -1 doesnt apply

[@AndreITA777]

Why dont use logaritm?

[@misterlau5246]

Apart from 1?

[@epikherolol8189]

Take log on both sides...

[@leoofficial527]

the line y=x is tangent to x^x at 1

[@TheMoogleKing93]

Dividing by X before considering if X = 0 (which isn't a valid solution) feels sloppy to me.

[@Nfsbelka]

0^0 is undefined

[@TheMoogleKing93]

@@Nfsbelka which he should have identified before dividing by x

[@seroujghazarian6343]

@@Nfsbelka no.

[@sohampinemath1086]

Just take log base x on both sides

[@hmwh4t]

Then you cannot get the -1 solution

[@sohampinemath1086]

@@hmwh4t ok

[@sathwikmadapati618]

USE LOG AND PROPERTIES ON BOTH SIDES to get 1,-1 and consider ln|x| . THEN U WILL GET ANSWER IN 2 SECS. Such simple 2 second questions you give for JEE ADVANCED INDIAN ASPIRANTS man😂😂. IF YOU TRULY WANT TO DO TOUGH MATH SUMS THEN TRY SOLVING JEE ADVANCED 2022 MATH PAPER IN 1 HOUR.

[@rv706]

Easy using convexity: 0) First of all, the proper domain of x^x is x>0 (You can do the case x = negative integer separately). 1) Observe that x=1 is a solution to x^x=x. 2) x^x is (at least) twice differentiable. Compute it's first derivative and evaluate at 1: you get 1, which means that y=x is the tangent to y=x^x at (1,1). 3) Now compute the second derivative of x^x and observe it's strictly positive for all (positive) x. So the function x^x is convex. 4) Convex functions lie above their tangent lines at every point (strictly above, for x different from the abscissa of tangency). Therefore, the graph of x^x lies above it's tangent line at x=1 (which is y=x). 5) Conclusion: x=1 is the _only_ solution to x^x=x.

[@Taric25]

Your conclusion is false. 0⁰=0, 1¹=1 and (-1)^(-1)=-1. There are clearly at least three solutions. 0⁰ is normally indeterminate, but subtracting x from both sides and factoring clearly shows x=0 is a solution in this case, since (-x+x^x)=0=(-1+x^(-1+x))(x).

[@Bobby-tj1hq]

Dude solve this is one second by using exponent rules. If base is same, you can equate exponents. Given x^x=x^1, since both sides are single exponents with base x, x=1.

[@narkitikx9380]

u can take the ln on both sides..

[@amritjanardhanan]

Then you will have x*ln(x) = ln(x); subtract ln(x) from either side and factor and you will have ln(x)*(x-1) = 0, so either x-1 = 0 and x = 1, or ln(x) = 0 so x = e^0 = 1

[@user-kiribati]

Is X^x continuous function?

[@user-kiribati]

@Jory Folker then we must search solutions only for x>0.

[@timbomb374]

The power has to be 1 or -1 because that's the only way you are going to have a number to the power of something also be itself. And since the power is an X that means all the Xs are either 1 or -1.

[@jefftobi]

There's something I don't get. So you have transformed the ecuation into x^(x-1)=1 and found the roots based on some basic rules, but the 2nd one is kinda debatable. If a=-1 that means that a^b is -1^b=-1/(-1)^|b| if b0 then -1^b=1 only if b=2n. So theres a change between odd and even numbers for b depending on whether it is a positive or negative inter. So x=-1 doesn't really work. At least thats how i see it

[@OriginalSuschi]

Suppose x = 2*y, where y is some real number Then (2y)^(2y)=2y Taking the ln on both sides: ln((2y)^(2y))=ln(2y) which is y*ln((2y)^2) = 1/2 *ln((2y)^2) we get the first solution by dividing by the ln(…) assuming it is not equal to zero. then y=1/2 and thus x=1 The other solution is when ln is 0, that’s the case when the inside is equal to 1, so (2y)^2=1 Take the square root and divide by two and we get two solutions: y=+1/2 and y=-1/2 Which leads once again to x=1 but also to the „hidden“ solution x=-1

[@aiaioioi]

why would you substitute x=2y if you can just do all of that with x tho, since it's not equal to 0 x*ln(|x|) = ln(|x|) ln(|x|)(x-1)=0 ln(|x|)=0 |x|=1 x=±1 (second gives us x=1 anyways)

[@aiaioioi]

like, we usually take logarithms without the modules since most of the times we're dealing numbers equal or more than 0

[@pierreocg]

Sir, i want to talk about case "x equal zero" Could we say 0^0=0?

[@harcha3562]

No i think. 0^0 is either 1 or undefined. Depends on which part of math we are talking about

[@etb753]

0^0 is either undefined or 1, which is not a solution

[@oZqdiac]

0^0 is indeterminate, some people say it is 0, some people say it is 1.

[@anic1716]

@@oZqdiac 0^0 can't be zero. There's an open discussion if it's 1 (lim x→0 x^x = 1) or indeterminate but it's can't be 0.

[@abhijitmajhi5895]

Simply we can find answer by seeing question

[@sumansharma8816]

Look at this....... There is an exponential rule which says that... If:- a^b=a^c Then, b=c so... x^x=x It is also be written as... x^x=x^1 then according to the exponential rule..... x=1

[@chainsawmanfan3002]

May I know why she complicating this problem.

[@Joseph-we8xw]

Multiply both sides by x You'll get X^x+1=x^2 Therefore x+1=2 X=2-1=1

[@NotScotishYellow]

If all x’s need to be the same then x=1 because 1^1=1=x^x=x

[@iamthepersonwhoasked5639]

I thought x is 1 because something with a power still stays the same thing if the power is 1 or such and since x is 1 the other x must also be 1

[@cantcode023]

why is x^x ÷ x = 1??

[@shadowzangoose]

If you divide both sides by x, that's what happens, since x ÷ x = 1 for all non-zero x

[@mr.cheese4169]

Answer:±1

[@starpawsy]

x=1 is an obvious solution by inspection.

[@presa.]

Logically no number elevated to itself can be equal to its base except for 1 and -1, this didn't even need an explanation since the equation speaks for itself.

[@vasiliykryuchkov7130]

I don't know math, could someone explain why we can't just say: a^x=a^1 then x=1?

[@LarryLasagna]

3:46 why can’t a=0? 0^0=1

[@viganadziratel1962]

There is a restriction for this equality x > 0, so the answer -1 is not suitable. If you plot x^x, you will see only one solution

[@mathwindow]

LHS = (-1)^(-1) = 1/(-1)^1 = 1/(-1) = -1 = RHS

[@masteruser97]

What? 😂😂. Where do u see the restriction?? Wake up buddy

[@viganadziratel1962]

@@mathwindow You're just lucky that by not imposing restrictions on x, you got the correct equality. For example, if you forget about the constraints when calculating the equation logx(x)=x, you will get wrong solutions. Let's go back to the original equation. The function x^x is only defined for positive values ​​of x, so you must impose the condition x > 0 in this equation

[@novidsonmychanneljustcomme5753]

@@masteruser97 Try x^x for x=-1/2 and you will see why it makes sense to define this kind of restriction, at least as long as you're dealing with a real-valued function. Of course you get real values for certain negative x inputs like negative integers, but not only (try x=-1/3 for example). Nevertheless it makes no sense to plot this function for negative x in the classical way because it is highly discontinuous there. There is a (in my eyes) wonderful video explaining all this in detail and also showing ways to get access to the understanding how functions like these can be imagined if you accept the introduction of complex numbers: kzbin.info/www/bejne/lZ3FYnSurL1_l68 (The part about x^x starts at 7:23 but I recommend to watch the whole video to see all contexts.)

[@riggsmarkham922]

@@novidsonmychanneljustcomme5753 who cares if it’s highly discontinuous there? Doesn’t make it invalid.

[@nishadvispute125]

taking log on both sides.... log x^x = log x x log x = log x x= log x / log x so x=1

[@awesomesky9492]

Taking log both side and cut log(x) to log(x) ........ x=1

[@Ulissescars]

This is 1, I don't know if there are other solutions, though.

[@lashavardidze2296]

-1

[@SharoPisanchov]

±1?

[@ACheateryearsago]

@@lashavardidze2296 true

[@NuGeera]

-1

[@Ulissescars]

I completely forgot -1 lol

[@tsomakilentswana4899]

The answer is plus and minus one by inspection

[@Nacho_Meter_Stick]

i have a math problem to share. turn (x*sqrt(4-x^2) - pi)/(2x^2 - 4) into (x^[something2])/[something] [something] cannot contain any variables [something2] also cannot contain any variables

[@ygalel]

Xlnx=lnx so x= 1 or e But e is extraneous so 1 is the only solution

[@francishunt562]

Fairly obvious without having to do the working out

[@oZqdiac]

Easy solutions, x = ±1. Some people will also say x = 0 as another solution but that is wrong since 0^0 is indeterminate

[@Heregoesnuttin]

Without skipping to the end, I'ma say "1".

[@oZqdiac]

Correct but also consider -1

[@mappingsworld]

X to power of x would be 1 to power of 1. 1(to power of 1) is basically 1 It will not work by anything else other than 1… 2(to power of 2) would equal 4, so NO 3(to power of 3), would equal 9, so NO I think u know what I am getting at

[@mappingsworld]

1 x 1 x 1 x 1… *I T. W I L L. A L W A Y S. B E. O N E*

[@hoshire_AOK4819]

x=+-1

[@pranjalrathore4804]

take log both sides and answer coming out is 1

[@miegorengq]

I want to ask something, why X^x=x^1 Its same with x^x÷x=1 ? Example, 2^4=2^4, Its not same with 2^4:2=4, 2^4=16, 16÷2=8, 8 not same with 4, so why x^x=x^1 Its same with x^x÷x=1 ?, Please aswer me, Im not understand

[@shoryaagarwal677]

How I did this was, x^x = x, so x = x^1/x, this implies 1/x = x, so x^2 = 1, therefore x = +1 or -1

[@Nikioko]

x^x = x x • x^(x-1) = x x^(x-1) = 1; x ≠ 0 logₓ(x - 1) = 0 lg(x - 1) / lg(x) = 0 lg(x - 1) = 0 x - 1 = 0 x = 1. But -1 is a solution as well, which I lost when I introduced the log.

[@novidsonmychanneljustcomme5753]

Your final solution is correct of course, but I don't understand what you're doing to get from step 3 to 4 as well as from 6 to 7... 🤔

[@Nikioko]

@@novidsonmychanneljustcomme5753 log 1 = 0 for any base.

[@novidsonmychanneljustcomme5753]

@@Nikioko You're right with what you're saying here but your algebraic steps are not plausible though. Step 3 to 4: If I got you right, you take log with base x on both sides. Indeed this makes 1 become 0, but what about the left side? You would get log_x(x^(x-1))=(x-1)*log_x(x)=x-1 which is not equivalent to what you wrote in step 4. And from lg(x-1)=0 you would get x-1=1, not =0 as you wrote from step 6 to 7. Nevertheless you get the right solution because these 2 algebraic mistakes "annihilate" each other in the end. ;) (Since you get to x-1=0 which is what results if you do step 3/4 correctly as I described above.)

[@Nikioko]

@@novidsonmychanneljustcomme5753 Of course it is equivalent. Logarithm of a number to a certain base is equal to logarithm of that number to any base divided by logarithm of the original base to the new base. That is one of the logarithmic rules, the one which lets you convert bases. So, logₓy = lgx/lgy = lnx/lny = log₂x / log₂y = ... As for the rest: I had about 30 seconds to solve this problem, I didn't check everything. But logₓ1 = 0 for any x ∈ ℝ⁺\{1} and vice versa.

[@novidsonmychanneljustcomme5753]

@@Nikioko That's right, but this is not what I meant, I was not referring to step 4 to 5. Maybe I expressed myself imprecisely - I should rather say "line" instead of "step" (which you maybe interpreted as respective mathematical transformation between the lines). So once again: Your transformation from *line* 3 to 4 is not correct. From x^(x-1)=1 it follows (after taking log_x on both sides) x-1=0 and NOT log_x(x-1)=0. And from *line* 6 to 7 there is also one mistake: You wrote lg(x-1)=0 x-1=0, but actually it would be lg(x-1)=0 x-1=1. As said: Your final solution is correct, but preceded by 2 algebraic transformation mistakes.

[@pendant_animations8757]

It would have been very simple if you would have added just one more step. x^x ÷ x = 1 => x^x-1 = 1 => x^x-1 = x^0 => x-1 = 0 => x = 1

[@kai1523]

Quite simple actually.

[@MnO_Brothers]

😏X^x=X The Xs are on the both sides of the equation. So the powers of both sides are also equal. So we can take X=+-1 as the solution easily without any hesitation👍

[@MikeN1811]

Автор, постоянно делает в своих роликах ошибку. Показательная функция y= a^x, для вещественных х, определена только для а>0. Значит это уравнение имеет только одно решение x=1.

[@TheFlairRick]

Can you try x^x= 2x, x^x=-x, and x^x=x+2?

[@joungi6135]

Nope

[@Zzz_Gamer]

I don't understand, isn't -1^(-1) = 1/(-1)? In that case x != -1.

[@aiaioioi]

x^x = (-1)^(-1) = 1/(-1) = -1 x= -1

[@diegocabrales]

It's not -1^(-1) but (-1)^(-1). If you do -1^(-1), the -1 exponent only affects to 1. Furthermore, (-1)^(-1) = 1/(-1) = -1, so it is a solution of the equation x^x = x.

[@rengokukyojuro1384]

x is 1 by observation

[@boeubanks7507]

The simplest way to solve this is too just equate the exponents. Simple logic, if the bases are identical, then the exponents have to have identical values to satisfy the equation. Therefore, x equals 1. So, the issue with negative one is a little trickier. If you say x = -1 then the equation, as written isn't really true because you have -1 ^-1 = -1. Those aren't identical equations. However, the rule for negative exponents is that you reduce them by changing them to the positive recripacle. So -1^-1 becomes -1^1. This is identical to the opposite side of the equation. However, this only works for -1 and 1 because the recripicale of 1 is 1. She got the right answer but, I still am not sure she did her math right. Something looked off about the way tried to address the negative root. Maybe someone else can explain what she did better.

[@brianpso]

What is this rule that I never heard of? Why can you simply delete the minus from the exponent?

[@boeubanks7507]

@@brianpso you invert the exponent (2 becomes 1/2, etc) and remove the negative sign. What your are essentially saying is that negative exponents are another way of writing positive roots. There are proofs out there if you want to look them up. Here is one random one I found: kzbin.info/www/bejne/qX7bc4malp2ErZo

[@brianpso]

@@boeubanks7507 My god yeah I remember this now! The example with 2 did the trick. Lol I feel so stupid for not remembering something so basic. Thanks man

[@predoshabido7504]

How can you prove that these are the only cases? Btw she included -1 because for example (-1)²=1, if you have any even exponent to -1 you get 1: given n as an integer (-1)^(2n) = [(-1)^n]^2 (-1)^n is either 1 or -1, but regardless, [(-1)^n]^2 = 1

[@mathserreurs2479]

Try x^(x+2)=x^(1-x) You have to get: 0,-1/2, 1 as solutions but not -1 because x+2-(1-x)=-1 not even. Other examples in: kzbin.info/www/bejne/l3illmyFl9afraM

[@hook-x6f]

I thought that was trick question because one popped in my head instantly but I am not really that good at math so I think ahh trick question. There must be another answer. Negative one also answer say hot stuff math window girl. I give thumbs up.

[@ruchikarfacts7380]

Olympiad questions 2^pq - 2^rs = 32; where p,q,r,s belongs to natural number then Possible values of p + q + r + s = ? Solution:- kzbin.info/www/bejne/mIO7i2Vjn8aCbtU

[@AlexeyEvpalov]

Спасибо.

[@robertlupa8273]

*[wrote this before watching]* You divide both sides by x. Right becomes x/x which is 1. (x^x)/x affects the power, which turns into x^(x-1). We get x^(x-1) = 1. Now I don't know how to explain this with equations or something, but logically, x has to be equal to 1, because 1^(1-1) is 1^0, which is 1. For every other number, you'd get for example 2^(2-1), which is 2^1, which equals 2, which is _not_ equal to 1. Same with every other number, positive or negative, or zero (which we excluded anyway earlier). Every number except 1. Is that the correct answer? 😌

[@defeatist3463]

No x can be -1 as well Booooo Jk nice

[@harleyjs4565]

If x^x / x^1 = 1 then 1 * x^1 = x^1 so x^x = x^1 that means x = 1

[@kartikschannel5297]

x=1 by observation

[@adgf1x]

x=Log of x base x=>x=1

[@trungoan7572]

x^x = x xlnx = lnx lnx(x-1) = 0 lnx = 0 or x-1= 0 => x=1. Anything wrong? Where's x=-1 with this way.

[@megamentebr7716]

Solution for the thumbnail equation: x = 1

[@charlesbromberick4247]

nice job

[@lool8421]

when in doubt, let x=1