An Exponential Equation | 27^x=1/x

[@creounity]

1/3 is easily guessable, and since there's just one intersection point, there are no more solutions.

[@mcwulf25]

The way to get full marks in this question is to prove your last statement.

[@seroujghazarian6343]

@@mcwulf25 why? xe^x is bijective for x>=0

[@crazycat1503]

@@mcwulf25 27^x is increasing and 1/x is decreasing

[@バナな-j5m]

If x≦0, x27^x≦0.This means that x27^x≠1. So,We don’t have to find negative x. x27^x is monotonically increasing function where x>0. So,x such that meets x27^x=1 is uniquely determined.

[@think_logically_]

We have x·(27^x)=1. Since the left-hand side is monotonically increasing, the equation may have only one solution. Rewrite the equation as x·3^(3x) = 1, then multiply by 3: (3x) · 3^(3x) = 3. Substitution y=3x gives y·3^y=3 or y·3^(y-1)=1, where we can easily guess that y=1, so x = y/3 = 1/3.

[@sudhakumari6415]

There is no need of doing so long as follows ---- 27^x=1/x 27=(1/x)^(1/x) 3^3=(1/x)^(1/x) Hence both RHS is equal to LHS-- THEREFORE,x=1/3 That's how I made. Like if u agreee

[@SyberMath]

Sure! 🙂

[@cablethelarryguy]

I like how rigorously you do these. Keep up the good work.

[@SyberMath]

Will do! Thank you 💖

[@ytlongbeach]

all that rigor is swell and all. however, i just guessed 1/3. done. lol

[@pcklop]

We can find general solutions using the lambert w function, where xe^x=y is solved by w(y)=x. The equation A^x=1/x has solutions x=w(ln(A))/ln(A). For the particular value A=27, plugging in gives x=1/3.

[@HoSza1]

My calculator doesn't have a Lambert W button, does yours? So how do you plug A into it?

[@leonardobarrera2816]

I like the w lambert function!!!

[@mcwulf25]

1/3 is easy guess and check. Is it enough to say that the two functions, call them f and g, are equal when f(1/3) = g(1/3). And f is monotonically increasing while g is monotonically decreasing for x>0? ie f < g if x < 1/3 and vv. For x 0 but g < 0 so no intersections.

[@srividhyamoorthy761]

dayum i have developed a lot of equation solving skills , i could figure out the answers to most of your videos now

[@BOBPERIO2]

Very nice video! Merry Christmas to you and your family. 🎅

[@SyberMath]

Thank you! Merry Christmas and Happy Holidays!!! 💖🎉🥳

[@scottleung9587]

I knew right away x=1/3 after a first glance.

[@alextang4688]

I had learnt this "special" method from you channel that I had not learnt at school. This is the one you always use and seems workable for some equations. It is nice! Merry Xmas!!!🎅🎅🎅🎅🎅🎅

[@SyberMath]

Excellent! Merry Xmas, Alex!!! 🥰

[@vashon100]

learned is better than learnt

[@moshyroth]

This will work for the general case. 4^x=1/x (x=1/2), 256^x=1/x (x=1/4), 3125^x=1/x (x=1/5). In general, equations in the form n^n^x=1/x will solve to 1/n, as n^n^x=n^n^(1/n)=1/x=n.

[@pNsB]

It seems to be possible to solve this with the Lambert W function. That was my approach and I got an answer, although evaluating it is a pain without googling it: I got W(ln(27)) / ln(27), which is actually approximately 1/3.

[@VolksdeutscheSS]

I did it by Lambert Method as well--successfully.

[@tyronekim3506]

That was a nice explanation. Thank you.

[@SyberMath]

Np. Thank you! 😍

[@ГеоргийПлодущев-с2н]

Преровняем обе части к y, имеем y=27^x y=1/x, график первой функции- экспонента , второй -гипербола, у них может быть только одна точка пересечения, следовательно, если мы найдём корень -он будет единственным. Это число 3 Ответ:3

[@homoexhumo]

We have 27^x = 1/x => 27=(1/x)^(1/x) let y = 1/x => 27 = y^y Now the solution is obvious, but we could find it by applying the recursive relationship: y = 1, y = (27^(1/y)+y)/2 which successively gives: y = 14 y = 7.6 y = 4.57 y = 3.31 y = 3.008 y = 2.999 ... y = 3 Therefore x = 1/y = 1/3

[@academiabartender6304]

You didn’t have to explore t^t function for monotonicity. After guessing the answer you could’ve just said that 27^x is increasing and 1/x decreasing, that’s enough

[@qwang3118]

Let y = 3x. Then y*(3^y) = 3. y = 1 is an easy solution. Since y*(3^y) is strictly increase in y > 0, y = 1 is the unique solution. so x = 1/3 is the unique solution.

[@Vinnan_Thamizh]

I found the answer in a few seconds after looking at it.

[@davidbrown8763]

So did I

[@tayserbinjafor1569]

Can't you do the Lambert W function yet? This problem can be solved by using the Lambert W function.

[@namoplayz5408]

You can use the lambient W function

[@jim2376]

x = 1/3 by inspection. Cross multiple. x(27^(1/3)) = 1. (1/3)(3) = 1.

[@limitless1035]

You can also use Lambert W function to solve this equation

[@theyousefkhan]

I was thinking that, but how do you apply it for 27? I thought it returned x where the expression is xe^x, not xa^x with a=27 for example

[@limitless1035]

@@theyousefkhan 27=e^ln27 You'll get x×e^(xln27) Then you just multiply by ln 27 to both sides xln27×e^(xln27)=ln27 Then use Lambert W function

[@SyberMath]

Too complicated! 😲😜😁🤣

[@RedAngel11993]

This is the faster way: after logaritmization, we get: 3x ln3=-lnx; 3ln3=-lnx/x; it follows x=1/3.

[@SyberMath]

How come I never thought of that? 😲😁

[@МаксимАндреев-щ7б]

x=27^(-x)>0, x increases, 27^(-x) decreases, x=1/3 is a solution -> this is unique solution

[@michaelrubin8194]

what was the point of analysis of the derivative from negative infinity to 1/e (in the table), if it is defined only for t>0???

[@SyberMath]

I don’t know 😜

[@JohnRandomness105]

Unfortunately, the only way I know of to solve this equation is by inspection or trial and error: 27^⅓ = 3 = 1/⅓. Also, for positive x, 27^x increases while 1/x decreases. If x is negative, 27^x is positive while 1/x is negative. Therefore, the solution is unique: x = ⅓.

[@seroujghazarian6343]

*cough* Lambert W function

[@nancydelu4061]

What fun!

[@ermattia]

I enjoyed it a lot!

[@SyberMath]

Glad you enjoyed it!

[@nikolasbradley3608]

When I tried this out for myself, i got (W(ln27))/(ln27) which is about .303. I multiplied by ex, rewrite in terms of e to the power, multiples by ln27 on both sides and this used Lambert, and divided over the rest, what is this value coming from?

[@SyberMath]

Lambert W sucks!!! 😜🤣

[@fullfungo]

Did you use the principal value of W? (it’s a multivalued function)

[@odysseasv7734]

Nice one!

[@SyberMath]

Thanks!

[@walterufsc]

All the analysis using calculus and graphing could have been done directly on the original functions of x, without having to do the more complicated analysis on the auxiliary variable t.

[@SyberMath]

Sure

[@mathisblanchot]

We can use W Lambert function

[@SyberMath]

😲😜😁

[@stevenlord7793]

after viewing and reading the comments I did not see a way to solve a more general situation where 27 is replaced by C, a positive constant. Numerical methods? Or is this where Lambert W comes in?

[@SyberMath]

yes

[@stevenlord7793]

@@SyberMath So then, maybe say how? Or perhaps it is too complicated? Without a general method, the the example here is contrived to be solvable by inspection.

[@hamitkoca1110]

excellent

[@GreenMeansGOF]

x = W(ln(27))/ln(27). The principle branch gives us 1/3.

[@peterruf1462]

Lambert w function is boring. It's like googleing the answer

[@angelmendez-rivera351]

@@peterruf1462 It isn't Googling the answer any more than saying x = sqrt(2) or x = -sqrt(2) for x^2 = 2 is Googling the answer.

[@peterruf1462]

@@angelmendez-rivera351 I generally agree with you but in this case you could simplify it to xln(27)e^xln(27)=ln(27)=ln(3)e^ln(3) ln(3)=xln(27) x=1/3 No "real" need for the w Lambert function. It is more like writing x=sqrt(4) as solution for x^2=4 instead of +/- 2

[@angelmendez-rivera351]

@@peterruf1462 That argument doesn't make sense. It seems to me that your issue here isn't with the Lambert W function, but with the fact that the answer hasn't been simplified. That's what the example you gave shows. The problem with writing sqrt(4) isn't that it uses sqrt, but that it isn't simplified.

[@peterruf1462]

@@angelmendez-rivera351 that is exactly my problem. Using the Lambert w function when it isn't "really" necessary. Oh well I wrote it in terms of the Lambert function let me just start up wolfram alpha to look at the principal branch.

[@gelbkehlchen]

Solution: 27^x = 1/x |*x≠0 ⟹ x*27^x = 1 = 1/3*27^(1/3) |the same operations are done with x on the left side of the equation and with 1/3 on the right side of the equation. Therefore must be: x = 1/3

[@legendeca5535]

If we simplified 27^x to 3^3x and cross multiplied to get 3x^3x =1^1 since the exponents are the same we could just change this into the equation 3x=1 which is x=1/3

[@RebornKaotic]

No you can only do that if its same base

[@sergeyshchelkunov5762]

Nice. Really detailed and thorough explanation how to solve this kind of problems.

[@SyberMath]

Glad to hear that!

[@gapplegames1604]

so easy. once you get (1/x)^(1/x)= 27, well, what raised to the power of itself equals 27? three. 1/x = 3, x = 1/3

[@individual1st648]

i just solve it by reasoning that since 27^x obviously is supposed to increase whereas 1/x doesnt, then it could be inferred that x must be a fraction and knowing cube roots the answer was evident

[@fireproyect9990]

The Lambert W function: 🤡

[@SyberMath]

😱😁🤪

[@negulescuanca8651]

The equation can not have negative solutions, obviously, The first member is an increasing function and the second a decreasing one, so the solution is unique and can be found immediatelly. Why so much trouble using derivatives?

[@explainingphysicsandmathematic]

Don't we need to check that there isn't the graph on -t axis for t^t...?

[@SyberMath]

Not well defined

[@joeldick6871]

Tea time for two!

[@SyberMath]

😄

[@beirirangu]

27^⅓=∛27=3 and ⅟⅓=3

[@Angels_i3]

Lambert W is a not nice looking but it gives a solution easily

[@pavlikkk101]

Solved in 1 second after one look :)

[@rogerkearns8094]

27: what's the first thing that anyone who loves numbers knows about it? Now, go figure.

[@user-pr6ed3ri2k]

t=1/3???

[@ftorum19]

I was sloving it for 5 sec. And I have gotten right Answer

[@martinjoster3282]

Guessed it before coming down here lol

[@levskomorovsky1762]

Please tell me if my decision is correct? х27^х = 1 х(3^3)^х =1 х3^3х = 1 х3^х3 = 1^1 Х3 = 1 х = 1/3

[@DayneTreader]

Using a perfect cube was too easy.

[@Kualinar]

Found 1/3 rather quickly. Just looking at the equation, it's obvious that X MUST be fractional and less than 1. Then, we know that 27 = 3^3. So, 27^(1/3) = 1/ (1/3) = 3

[@02lyn73]

27=3^3 1/x=3 X=1/3

[@hazalouldi7130]

on peut le resoudre zvec la fonction lambert w(3ln3)=ln(1÷x)

[@deltalima6703]

Um, what about x

[@mickanton7285]

simple question but you made it very complicated

[@SyberMath]

That’s me! 🤪

[@mickanton7285]

@@SyberMath Enjoy the block

[@shorts_for_shortcuts]

Can't you use logbase 3 to t^t and get t = 3

[@vitalsbat2310]

Lambert W function

[@salvatoremanzo]

X=1/3

[@broytingaravsol]

x=1/3

[@moeberry8226]

Too easy Syber this wasn’t even a challenge.

[@scottleung9587]

Well, it was easy to guess and check but hard to go thru all the rigorous analysis.

[@SyberMath]

You’re too good! 🥸😁

[@moeberry8226]

@@SyberMath lol

[@moeberry8226]

@@scottleung9587 there’s no deep analysis needed, Syber made it more complicated than it needs to be. 27^x is an increasing function and 1/x is a decreasing function for all x. So max there is 1 solution game over.

[@user-lu6yg3vk9z]

@@SyberMath 27^x is an increasing function and 1/x is decreasing so therefore there is only one solution.

[@neuralwarp]

You didn't prove that f(t) is decreasing on [0,1/e)

[@SyberMath]

The table shows it

[@diegocabrales]

@@SyberMath You took some close value at the right-hand side of 1/e and saw that f(t) > 0 for that value. However, you only said that for some close value at the left-hand side of 1/e, f(t) < 0. I think that was what neuralwarp was trying to say.

[@ganda3454]

3^3x=x^-1

[@joyceenjoy-reading7914]

guess for x =1/3 within 10 seconds

[@岸辺緑]

直観で 1/3に気づければ、あとは他の実解がないのを詰めればよいが そうでなければ苦戦

[@explainingphysicsandmathematic]

Actually I checked google why there isn't the graph for minus side sir ?

[@SyberMath]

Not well defined

[@julianbruns7459]

f(t) = t^t(1+ln(t)) and the logarithm is only defined for positive numbers when talking about real numbers.

[@Roq-stone]

I can’t believe you threatened the lives of two birds and your video didn’t get taken down!! 😅

[@SyberMath]

Oh, no! 😳🤫😁😂

[@yoshinaokobayashi1557]

27=3^3 then x=1/3

[@jensjacobs9050]

in Holland we say : use your head. It took me 3 seconds with my head

[@famtik_arshi7547]

27^(1÷27)= x^(1÷x) x=? can you pls solve.

[@-basicmaths862]

Put 3x=t, Then x=t/3 &1/x=3/t then solving this equation We get t=1 & x=1/3

[@tryingtomakeanamebelike7245]

Imagine not knowing your roots and just being able to say 1/3 without doing math

[@SyberMath]

😀

[@giuseppemalaguti435]

x=W(ln27) /ln27=1/3

[@yackawaytube]

1/3 easy. x can't be > 1. 0 is too small. 27 to any power < 1 gives an irrational number except for 1/3 since 27 is 3^3. No fraction can match an irrational number.

[@matthewmcdaid7962]

X = 1/3. 27 to the 1/3 power (third root of 27) is 3, which is 1 / 1/3. QED

[@arturovinassalazar]

So me: x=W(ln(27))/ln(27) xDDD ¿There isnt a formula for W function?

[@novemtrigintillionaire7684]

bro using calculus while I found the answer 1/3 in 24 seconds

[@SyberMath]

😁

[@omurgokcinaar2501]

I was doubting if you were Turkish or not because of your accent. But then you said you liked tea and now I know for sure that you are Turkish hhhhhh

[@SyberMath]

😁😍😊😉

[@geoffreyparfitt7003]

What index fits well with a cube number?.. One third?...Oh yes that works.

[@SyberMath]

😄

[@justinnitoi3227]

I used lambert W

[@SyberMath]

😱😜

[@benitoregaloncastro3725]

X=1/3.

[@gregoriopisconte4628]

Demasiado fácil x igual , 1/3

[@ammuvilambil8032]

1/3

[@phrtao]

The first thing you notice is that 0 < x < 1; then you notice that 27 is actually a cube value. After that the numbers just fall into place.

[@user-nm6yy1uq3d]

X=1/3 easy

[@nozdref1567]

If you can't get rid of something, let's (ln) both sides!

[@kropek1754]

I mean I knew it by simply taking a look xD

[@SyberMath]

👍

[@폭풍간지남-o8v]

As a korean this equation can solve by 7yo korean kid

[@SyberMath]

Wow!

[@barakathaider6333]

👍

[@davidbrown8763]

No need for much of this. I solved this in 3 lines, without functional notation, logs or graphs - just simple algebra. After establishing that t = 3, you could have simply concluded that x=1/t =1/3.

[@SyberMath]

Nice

[@davidbrown1614]

@@SyberMath Thank you.

[@무명-q2w]

삼분의 일 ㅋㅋ

[@faithieTAKEN]

insanin aksaninin olmasi ne guzel sey ya

[@SyberMath]

😁

[@johnchurch2632]

Way too much explanation for a simple equation. with no lengthy explanation. Some people go overboard with their thinking, no wonder some people hate Math (Retired Math teacher here, LOL)

[@jwang734]

Overdone, 1/3 is easily seen.

[@matteomitrano4128]

I can't understand why you should complicate your life with all this jazz and nonsense. 27=(1/x)^(1/x) Let 1/x=n 3³=nⁿ n=3 n=1/x=3 x=1/3 Simple and without too many frills

[@langleywallingford260]

Umm... x = 1/3... pretty easy to figure out...

[@langleywallingford260]

Why did he need to give such a long, mind bogglingly, complicated explanation with endless digressions when the answer is obviously one-third??? 🙄🙄🙄