The easier solution is to do substitution x --> 2^m. Then, we get 2^m = 32m, or m = 2^(m-5), hence m is also a power of two (if we are concerned only with the integer solutions). The smallest value m=8 satisfies the equation. hence m=8, x=2^8 = 256.
2^x = x^32 ln2 2^x = ln2 x^32 x ln2 2 = 32 ln2 x x = 32 ln2 x x / (ln2 x) = 32 x x / (ln2 x) < 0 -> x must larger than 1 x = 256 -> 256 / (ln2 256) = 32
@@noname-ed2un He starts from the assumption that X will be a power of 2. This is a good start for finding a solution, since it reduces the complexity of the problem, because both sides are now powers of two. Set x = 2^m 2^(2^m) = (2^m) ^32 2^(2^m) = 2^32m ------ power of power --> multiply exponents 2^m = 32m ---- equal powers --> equal exponents Now this is still not trivial but left side is a power of 2 and right side has 32, which is also a power of 2. Hence m itself must be a power of 2. Set m = 2^k 2^(2^k) = 32 * 2^k 2^(2^k) = 2^5 * 2^k 2^(2^k) = 2^(5+k) 2^k = 5+k Now we're down to small numbers and we can do a quick check for k = 1, 2, 3 ... to find 2^3 = 8 = 5+3 So m = 8 and x = 2^8 = 256 This was a bit lengthy but hopefully helpful.
Your answer is right but when you analize the graphic of 2^x and x^32 the graphic of both functions intersects at three points: x ≈ -0.979, x ≈ 1.022 AND x = 256. For those first two solutions I couldn't find an analytical way to calculate their exact value. I know they are irrational but I couldn't find a way to express them with roots or even something with the number e.
Unless x was restricted to integers, i would rate this answer as incomplete. Although there is only 1 integer solution, there are 3 real solutions (and an infinite amount of complex solutions). Did i miss the part where it was specified what set x belongs to?
@@ayugaming3047 First of all I thought the answer to this question could be expressed as 2^t. Then I substituted and compared the exponential part: 2^t=32t and 2^t is monotonically increasing, so there is one solution. So t=8. Sorry for my poor English
It's alright. Solving for irrational roots requires application of the Lambert W function or using numerical methods, so they can't be found by elementary algebraic transformations anyway.
In addition to the solution x=256, the equation also has a real solution between -1 and 0. With a problem like this, it would be nice to be specific whether to find all real solutions, or to find only solutions that are positive integers.
Иван, каким образом были найдены два других корня (помимо 256)? При проверке корня 1,022393 число сходится только до пятого разряда после точки/запятой, далее отличается
Привет Антон! Я применил функцию Ламберта W(ze^z)=z 2^x=x^32 Извлекаем корень из 32 x=+/- 2^(x/32) a=e^(Lna) x=+/-e^[Ln(2^(x/32))]=+/-e^[(xLn2)/32] xe^[(-xLn2)/32]=+/-1 Домножим обе части на -(Ln2)/32 [(-xLn2)/32]e^[(-xLn2)/32]=+/-(Ln2)/32 Возьмём функцию Ламберта от обеих частей W{[(-x(Ln2)/32]e^[(-xLn2)/32]}= W[+/-(Ln2)/32] -x(Ln2)/32=W[+/-(Ln2)/32] x=-32W[+/-(Ln2)/32]/Ln2 При помощи калькулятора для функции Ламберта находим W[+(Ln2)/32]=0,02120634 Ограничимся точностью в 7 значащих цифр x1=-(0,02120634 Ln2)/32=/0,9790196 W[-(Ln2)/32]=-0,02214590 x2=-(-0,02214590 Ln2)/32=1,02239294 Второе значение W[-(Ln2)/32]=-5,545177 x3=-(-5,545177 Ln2)/32=256 Последний корень можно получить точно. Преобразуем выражение -(Ln2)/32 к виду ze^z -(Ln2)/32=-(2^3Ln2)/2^3*32=-(8Ln2)/2^8=-(8Ln2)/e^(8Ln2)=-(8Ln2)*e^(-8Ln2) W[-(8Ln2)*e^(-8Ln2)]=-8Ln2 x3=-(-8Ln2)*32/Ln2=256
2^x= x^32 Posons x=2^n L èquatiion devient 2^(2^n)=(2^n)^32 Donc 2^n=32n en identifiant les exposants Posons n=2^a L’équation devient 2^(2^a}=(2^5)*(2^a) Donc 2^(2^a}=2^(5+a) Donc 2^a=5+a en identifiant les exposants Or 2^3=8=5+a Donc a=3 Donc n=2^3=8 Donc x=2^8=256 Donc 2^256=256^32 On vérifie : 2^256=(2^8)*32=2^(8*32)=2^256
@@TirthParmar-cw4dc There are infinite complex solutions. Complex numbers are defined by their absolute value and their argument, which is just the angle in the complex plane. And you can add 2πz (z∈Z) to the argument, then you get the same complex number again with a different angle. (2π = 1 full circle) Example: -1 = e^(iπ) = e^(i*(π+2π) = e^(3iπ) = e^(5iπ) = e^(7iπ) = ...
It is bad enough that only one solution is found. What is worse, is that uniqueness is not even discussed. Why so many people like to lecture on mathematics without first learning the basics of ``mathematical culture".
You spend an enormous amount of time finding the integer solution, while you fail to discuss the non-integer solutions (or complex solutions). Math is not just endless steps of algebra to arrive at prior knowledge. Math is about elegance in finding the solution and about rigor in figuring out if there are more solutions (depending on the solution space, which you need to define first).
Whoa hold up. 2*1/32= 2/32 or 1/16. When multiplying a fraction by an integer, the divisor stays the same or simplifies. You are only performing the multiplication on the numerator. 1/16=1/x. X=16. Why make this more difficult?
Hola cómo estás todo ustedes bueno día desde san Felipe de Puerto plata primera espalda de la restauración general Gregorio luperon machete carajo 🇩🇴⚔️☕
there is a faster and more accurate way: x > 0, obvious x cannot be non integer rational number because then LHS = irrational and RHS = rational. So, x > 1 and x is a natural number. Let x be 2 to the power of k. k > 0. So, 2 to the k = 32k. So, 2 to the (k-5) = k. Obvious solution: k = 8, so x = 256. Draw graphs for y = k and y = 2 to the (k-5). For k>8 they cannot meet again (show with differentiation). For 1
@@KyriZee > x > 0, obvious Why is it obvious? Both 2ˣ and x³² are defined for all x ∈ ℝ. > x cannot be non integer rational number But it can be irrational. > I have no idea how to solve for irrational. Using the Lambert W function, obviously. But the solution for an integer x is indeed accurate and neat.
Я применил функцию Ламберта W(ze^z)=z 2^x=x^32 Извлекаем корень из 32 x=+/- 2^(x/32) a=e^(Lna) x=+/-e^[Ln(2^(x/32))]=+/-e^[(xLn2)/32] xe^[(-xLn2)/32]=+/-1 Домножим обе части на -(Ln2)/32 [(-xLn2)/32]e^[(-xLn2)/32]=+/-(Ln2)/32 Возьмём функцию Ламберта от обеих частей W{[(-x(Ln2)/32]e^[(-xLn2)/32]}= W[+/-(Ln2)/32] -x(Ln2)/32=W[+/-(Ln2)/32] x=-32W[+/-(Ln2)/32]/Ln2 При помощи калькулятора для функции Ламберта находим W[+(Ln2)/32]=0,02120634 Ограничимся точностью в 7 значащих цифр x1=-(0,02120634 Ln2)/32=/0,9790196 W[-(Ln2)/32]=-0,02214590 x2=-(-0,02214590 Ln2)/32=1,02239294 Второе значение W[-(Ln2)/32]=-5,545177 x3=-(-5,545177 Ln2)/32=256 Последний корень можно получить точно. Преобразуем выражение -(Ln2)/32 к виду ze^z -(Ln2)/32=-(2^3Ln2)/2^3*32=-(8Ln2)/2^8=-(8Ln2)/e^(8Ln2)=-(8Ln2)*e^(-8Ln2) W[-(8Ln2)*e^(-8Ln2)]=-8Ln2 x3=-(-8Ln2)*32/Ln2=256
But at the end of your working; it doesn't make sense ;how took it( 256 the power of 1over 256 is equal x 1 over x ) how you equated But i appreciate you wow.... Can you say sth on that , what i said
The easier solution is to do substitution x --> 2^m. Then, we get 2^m = 32m, or m = 2^(m-5), hence m is also a power of two (if we are concerned only with the integer solutions). The smallest value m=8 satisfies the equation. hence m=8, x=2^8 = 256.
Can you explain this more. I don't understand
2^x = x^32
ln2 2^x = ln2 x^32
x ln2 2 = 32 ln2 x
x = 32 ln2 x
x / (ln2 x) = 32
x x / (ln2 x) < 0 -> x must larger than 1
x = 256 -> 256 / (ln2 256) = 32
@@noname-ed2un He starts from the assumption that X will be a power of 2. This is a good start for finding a solution, since it reduces the complexity of the problem, because both sides are now powers of two. Set x = 2^m
2^(2^m) = (2^m) ^32
2^(2^m) = 2^32m ------ power of power --> multiply exponents
2^m = 32m ---- equal powers --> equal exponents
Now this is still not trivial but left side is a power of 2 and right side has 32, which is also a power of 2. Hence m itself must be a power of 2. Set m = 2^k
2^(2^k) = 32 * 2^k
2^(2^k) = 2^5 * 2^k
2^(2^k) = 2^(5+k)
2^k = 5+k
Now we're down to small numbers and we can do a quick check for k = 1, 2, 3 ... to find 2^3 = 8 = 5+3
So m = 8 and x = 2^8 = 256
This was a bit lengthy but hopefully helpful.
DIFFICULT Question but you explain very easily
Your answer is right but when you analize the graphic of 2^x and x^32 the graphic of both functions intersects at three points: x ≈ -0.979, x ≈ 1.022 AND x = 256. For those first two solutions I couldn't find an analytical way to calculate their exact value. I know they are irrational but I couldn't find a way to express them with roots or even something with the number e.
can´t be solved analytically
@@PIANOJOE7why?
Also realized that the equation has 3 roots, can't find an analytical solution (
@@PIANOJOE7 it is possible! Take a look at the Lambert-W-Function. In Wolframalpha to be used as productlog(x).
@@adrianlautenschlaeger8578 Well, good hint. This function was unkown to me (or already forgoten 🙂...). So you´re right.
Very good-question and explanation
Hi, and thank you for your videos.
How can you be sure this solution, obtainend by identification (is that the word ?) is unique ?
Simply awesome 👌. I liked.
Nd subscribed too.
Those interested in maths will take interest in learning this 🎉🎉🎉🎉🎉🎉 Superb 🎉
It's really otherworldly! Fantastic!
I like your way is so amazing
Unless x was restricted to integers, i would rate this answer as incomplete. Although there is only 1 integer solution, there are 3 real solutions (and an infinite amount of complex solutions). Did i miss the part where it was specified what set x belongs to?
That's why I was thinking of using natural log and the Lambert W function
She was beautiful❤❤❤❤
xは多分2^tと書ける、指数部分比較して2^t=32tを解く→t=8からx=256
Could you please clarify my brother? The steps ...
@@ayugaming3047 First of all I thought the answer to this question could be expressed as 2^t. Then I substituted and compared the exponential part: 2^t=32t and 2^t is monotonically increasing, so there is one solution. So t=8. Sorry for my poor English
@@ああ-s5m8m your English is fine.
So, what calculations did you use after the step 2^x = 32t?
Did you just guess?
@@ayugaming3047 Mental arithmetic
@@ああ-s5m8m can you introduce that to me?
The other solution is: 1/e^lambertw(-(ln(2)/32))≈1.0223
But there should also be a negative real solution, since (−x)³² = x³² and 2ˣ = (1/2)⁻ˣ, so we get an equation (1/2)⁻ˣ = (−x)³² for −x ∈ ℝ.
@@allozovsky Functions 2^x and x^32 intersect only in two points.
You may plot them on the interval from −1.5 to 1.5 and find out that they indeed intersect at two non-integer points (and a third one at x = 256).
@@allozovsky yeah, my bad.
It's alright. Solving for irrational roots requires application of the Lambert W function or using numerical methods, so they can't be found by elementary algebraic transformations anyway.
Harika(perfect)👏🏻👏🏻👏🏻
Thank you very much 😊 ☺️ 🙏
Good excercise.
Very nice! I wondered where you were going but suddenly you arrived.
That's what she said
In addition to the solution x=256, the equation also has a real solution between -1 and 0. With a problem like this, it would be nice to be specific whether to find all real solutions, or to find only solutions that are positive integers.
Very good
Thanks 🙏❤️🙏
В этом уравнении имеется 3 корня
-0,97902
1,022393
256
Иван, каким образом были найдены два других корня (помимо 256)? При проверке корня 1,022393 число сходится только до пятого разряда после точки/запятой, далее отличается
Привет Антон!
Я применил функцию Ламберта
W(ze^z)=z
2^x=x^32
Извлекаем корень из 32
x=+/- 2^(x/32)
a=e^(Lna)
x=+/-e^[Ln(2^(x/32))]=+/-e^[(xLn2)/32]
xe^[(-xLn2)/32]=+/-1
Домножим обе части на -(Ln2)/32
[(-xLn2)/32]e^[(-xLn2)/32]=+/-(Ln2)/32
Возьмём функцию Ламберта от обеих частей
W{[(-x(Ln2)/32]e^[(-xLn2)/32]}=
W[+/-(Ln2)/32]
-x(Ln2)/32=W[+/-(Ln2)/32]
x=-32W[+/-(Ln2)/32]/Ln2
При помощи калькулятора для функции Ламберта находим
W[+(Ln2)/32]=0,02120634
Ограничимся точностью в 7 значащих цифр
x1=-(0,02120634 Ln2)/32=/0,9790196
W[-(Ln2)/32]=-0,02214590
x2=-(-0,02214590 Ln2)/32=1,02239294
Второе значение
W[-(Ln2)/32]=-5,545177
x3=-(-5,545177 Ln2)/32=256
Последний корень можно получить точно.
Преобразуем выражение -(Ln2)/32 к виду ze^z
-(Ln2)/32=-(2^3Ln2)/2^3*32=-(8Ln2)/2^8=-(8Ln2)/e^(8Ln2)=-(8Ln2)*e^(-8Ln2)
W[-(8Ln2)*e^(-8Ln2)]=-8Ln2
x3=-(-8Ln2)*32/Ln2=256
@@ivangorin1254 Иван, привет! Благодарю за разъяснение! Теперь понял :)
El problema tiene 3 soluciones: x= 1.02239, x= -0.97902
Also 1.0223
2^x= x^32
Posons x=2^n
L èquatiion devient 2^(2^n)=(2^n)^32
Donc 2^n=32n en identifiant les exposants
Posons n=2^a
L’équation devient 2^(2^a}=(2^5)*(2^a)
Donc 2^(2^a}=2^(5+a)
Donc 2^a=5+a en identifiant les exposants
Or 2^3=8=5+a
Donc a=3
Donc n=2^3=8
Donc x=2^8=256
Donc 2^256=256^32
On vérifie : 2^256=(2^8)*32=2^(8*32)=2^256
Smart solving ! Thanks.
Just last eq. 2^a = (a+5) how to obtain a=3 ?
Pourquoi avez vous mis x=2^n alors que vous ne connaissez pas sa valeur exacte? Expliquez moi s'il vous plaît
It was a lengthy solution. Rather it can be solved by taking natural log on both sides
very good. Thanks you!!!!!!!!
Wow that was a very entertaining proof!!
Thank you very much!
Bir ile iki arasında bir x daha olsa gerek onunda bulabilir misiniz?
GOOD IDEA
this answer is imcomplete. there are infinite amount of solutions.(x=256 is not the only one)
why infinite?
No bro only one 256 satisfied this equation
@@TirthParmar-cw4dc There are infinite complex solutions. Complex numbers are defined by their absolute value and their argument, which is just the angle in the complex plane. And you can add 2πz (z∈Z) to the argument, then you get the same complex number again with a different angle. (2π = 1 full circle)
Example:
-1 = e^(iπ) = e^(i*(π+2π) = e^(3iπ)
= e^(5iπ) = e^(7iπ) = ...
6
@@TirthParmar-cw4dc
3 real solutions:
x₁ = 1.0223929402057803206527516798494005683768365119132864517728278977...
x₂ = 256
x₃ = -0.979016934957784612322582550011650068748090048886011676265377083...
and many complex solutions
Tricky Question but you explain very well ❤.
What is the 3rd solution? It must be a small negative number
Yeah, it's a solution to (1/2)⁻ˣ = (−x)³²
Approximately -0.979.
@@alejomdp How did you get it? Numerically or evaluated via some tool?
@@allozovsky wolfram alpha, it's a calculator
Very nicely proceeding towards the target
Very good solution
Cette méthode pose le problème de la bijectivité de la fonction x^(1/x), non ?
just use the logarithm equation
It is impossible in this case
Wow very nice
Méthode intuitive.
On remarque que 2^8=256 et que 8*32=256
Du coup on suppose x=2^8
puis on vérifie : 2^256=(2^8)*32=2^(8*32)=2^256
It was a lengthy solution but wonderful
Your welcome
X. 2^x. X^32
0. 1. > 0
1. 2. > 1
2. 4.
Because 256
yeah aprox. 1.022, for that you have to use the lambert function (w). Also another answer would be aprox. -0.979
It is bad enough that only one solution is found. What is worse, is that uniqueness is not even discussed. Why so many people like to lecture on mathematics without first learning the basics of ``mathematical culture".
Three guess' I got it. I like to solve things by experience and guesses too.
Just looked back at the problem and I understand You, there are infinite solutions if you do not have use Integers, they did not specify it so...
Make your video. I wanna see your solutions
This needs lambert W function to find 3 real solutions (infinite complex ones).
Can the Lambert W-Function help us to find the solution to this equation?
Me encanta cómo dice "two".
Use the Lambert W function W(■*e^■) = ■
2^x = x^32
ln(2^x) = ln(x^32)
x*ln(2) = 32*ln|x| ===> two cases
1st case: x > 0
x*ln(2) = 32*ln(x)
ln(x)*x^(-1) = ln(2)/32
ln(x)*e^ln(x^(-1)) = ln(2)/32
ln(x)*e^(-ln(x)) = ln(2)/32
-ln(x)*e^(-ln(x)) = -ln(2)/32
W(-ln(x)*e^(-ln(x))) = W(-ln(2)/32)
-ln(x) = W(-ln(2)/32)
ln(x) = -W(-ln(2)/32)
x = e^(-W(-ln(2)/32)) ===> -1/e < -ln(2)/32 < 0 ===> 2 real solutions
x₁ = e^(-W₀(-ln(2)/32)) = 1.0223929402057803206527516798494005683768365119132864517728278977...
in WolframAlpha: e^(-productlog(0,-ln(2)/32))
x₂ = e^(-W₋₁(-ln(2)/32)) = 256
in WolframAlpha: e^(-productlog(-1,-ln(2)/32))
2nd case: x < 0
x*ln(2) = 32*ln(-x)
ln(-x)*x^(-1) = ln(2)/32
-ln(-x)*x^(-1) = -ln(2)/32
ln(-x)*(-x)^(-1) = -ln(2)/32
ln(-x)*e^ln((-x)^(-1)) = -ln(2)/32
ln(-x)*e^(-ln(-x)) = -ln(2)/32
-ln(-x)*e^(-ln(-x)) = ln(2)/32
W(-ln(-x)*e^(-ln(-x))) = W(ln(2)/32)
-ln(-x) = W(ln(2)/32)
ln(-x) = -W(ln(2)/32)
-x = e^(-W(ln(2)/32))
x = -e^(-W(ln(2)/32)) ===> ln(2)/32 > 0 ===> 1 real solution
x₃ = -e^(-W₀(ln(2)/32)) = -0.979016934957784612322582550011650068748090048886011676265377083...
in WolframAlpha: -e^(-productlog(0,ln(2)/32))
Thank you sir❤
Nice bro🎉
You spend an enormous amount of time finding the integer solution, while you fail to discuss the non-integer solutions (or complex solutions).
Math is not just endless steps of algebra to arrive at prior knowledge. Math is about elegance in finding the solution and about rigor in figuring out if there are more solutions (depending on the solution space, which you need to define first).
2^x=x^32, so xlog2=32logx, then x/logx=32/log2=2^5/log2=2^6/log(2^2)=2^7/log(2^4)=2^8/log(2^8), so x = 2^8.
Can i use the lambert W function
W Lambert is useable for this
@@nazmurrahmantamim6014 YES!
2^x = x^32
ln(2^x) = ln(x^32)
x*ln(2) = 32*ln|x| ===> two cases
1st case: x > 0
x*ln(2) = 32*ln(x)
ln(x)*x^(-1) = ln(2)/32
ln(x)*e^ln(x^(-1)) = ln(2)/32
ln(x)*e^(-ln(x)) = ln(2)/32
-ln(x)*e^(-ln(x)) = -ln(2)/32
W(-ln(x)*e^(-ln(x))) = W(-ln(2)/32)
-ln(x) = W(-ln(2)/32)
ln(x) = -W(-ln(2)/32)
x = e^(-W(-ln(2)/32)) ===> -1/e < -ln(2)/32 < 0 ===> 2 real solutions
x₁ = e^(-W₀(-ln(2)/32)) = 1.0223929402057803206527516798494005683768365119132864517728278977...
x₂ = e^(-W₋₁(-ln(2)/32)) = 256
2nd case: x < 0
x*ln(2) = 32*ln(-x)
ln(-x)*x^(-1) = ln(2)/32
-ln(-x)*x^(-1) = -ln(2)/32
ln(-x)*(-x)^(-1) = -ln(2)/32
ln(-x)*e^ln((-x)^(-1)) = -ln(2)/32
ln(-x)*e^(-ln(-x)) = -ln(2)/32
-ln(-x)*e^(-ln(-x)) = ln(2)/32
W(-ln(-x)*e^(-ln(-x))) = W(ln(2)/32)
-ln(-x) = W(ln(2)/32)
ln(-x) = -W(ln(2)/32)
-x = e^(-W(ln(2)/32))
x = -e^(-W(ln(2)/32)) ===> ln(2)/32 > 0 ===> 1 real solution
x₃ = -e^(-W₀(ln(2)/32)) = -0.979016934957784612322582550011650068748090048886011676265377083...
Интересное задание и его решение!
We have a^2=2^2 leading to 2 solutions which are -2 and 2. What make you believe that the fact we have a^(1/a)=x^(1/x) gives a unique solution a=x??
excellent
Thank you! Cheers!
2x = 32x
Log 32 / log 2 = 5
2^5 = 32
YOU TALK WELL SIR,
You can solve using the concept of logarithms
How I try to solve
By applying logarithm,
(Logx/Log3) = x/9.
Next step??
The process is good for written tests but it's too long for compitive exams.
Thonk you ❤❤❤❤❤
Très intéressant !
Thank you very much!
Whoa hold up. 2*1/32= 2/32 or 1/16. When multiplying a fraction by an integer, the divisor stays the same or simplifies. You are only performing the multiplication on the numerator.
1/16=1/x. X=16. Why make this more difficult?
Très intéressant
Thank you very much !
X=0 ?? If 0^x=1 or is that for 0!
Çözüm cok uzun malesef.🤷♀️ 2 nin kuvvetlerini dene, ×=2⁸=256 sağlıyor.
Thanks
Сочетание логики и метода научного тыка
Wow Chat GPT can not answer. but you can.
are u from India?
(1)^x/32=2^32 ×x...so x= 1/2^32....
Tricky
Finnally bro did it
Take two sheets of paper!
It's so complicated way to solve 😢
Hola cómo estás todo ustedes bueno día desde san Felipe de Puerto plata primera espalda de la restauración general Gregorio luperon machete carajo 🇩🇴⚔️☕
Azərbaycancan salamlar niqırr(jholooom)
Alqasımlıdan salamlar helom məllim😂😂😂😂
It is just a binary number problem. ((32)^2)/(2^2) = 256
2 *8 equal x power x 16 square.
Güzel yöntemmiş ama çok uğraştıcı
How did i get 1
👍🙏
the answer is 32x32=1024
2x2=4
1024/4=256
At 71, x is 2 or -2
can i know the brand name of the pen you used in this video
why are they called Arabic numbers?
👍
👍👍👍
looks like a binary problem, x guesses would be 1, 2, 8, 16, 32.....256! No algeabras here!
Not Bad like methode
Thank you very much!!!
Just take log both sides
Hello dear I am from Ethiopia and I don't understand logarithm so do u know any good channel to help me
Log without calculator is not possible in math Olympiad that's where IQ comes in
This is not complete solution, so there are two more solutions, so three in total !!!
There are three real solutions in total: one integer solution and two non-integer solutions (a positive and a negative around 1 by absolute value).
@@allozovsky I know, just I wrote incorrect. I fixed it.
She is a beautiful ❤❤❤❤
Ty
Just take log of both side
Это не решение, а подбор только одного из трех корней. Количество трех корней можно увидеть графически.
Или использoвaть функцию Ламберта W
2^x = x^32
ln(2^x) = ln(x^32)
x*ln(2) = 32*ln|x| ===> два случая
1°: x > 0
x*ln(2) = 32*ln(x)
ln(x)*x^(-1) = ln(2)/32
ln(x)*e^ln(x^(-1)) = ln(2)/32
ln(x)*e^(-ln(x)) = ln(2)/32
-ln(x)*e^(-ln(x)) = -ln(2)/32
W(-ln(x)*e^(-ln(x))) = W(-ln(2)/32)
-ln(x) = W(-ln(2)/32)
ln(x) = -W(-ln(2)/32)
x = e^(-W(-ln(2)/32)) ===> -1/e < -ln(2)/32 < 0 ===> два корня
x₁ = e^(-W₀(-ln(2)/32)) = 1.0223929402057803206527516798494005683768365119132864517728278977...
x₂ = e^(-W₋₁(-ln(2)/32)) = 256
2°: x < 0
x*ln(2) = 32*ln(-x)
ln(-x)*x^(-1) = ln(2)/32
-ln(-x)*x^(-1) = -ln(2)/32
ln(-x)*(-x)^(-1) = -ln(2)/32
ln(-x)*e^ln((-x)^(-1)) = -ln(2)/32
ln(-x)*e^(-ln(-x)) = -ln(2)/32
-ln(-x)*e^(-ln(-x)) = ln(2)/32
W(-ln(-x)*e^(-ln(-x))) = W(ln(2)/32)
-ln(-x) = W(ln(2)/32)
ln(-x) = -W(ln(2)/32)
-x = e^(-W(ln(2)/32))
x = -e^(-W(ln(2)/32)) ===> ln(2)/32 > 0 ===> один корень
x₃ = -e^(-W₀(ln(2)/32)) = -0.979016934957784612322582550011650068748090048886011676265377083...
×^+/×=2^1/32
=2^2/2^6=4^1/64
=4^2/128=16^128
=16^2/256=256^1/256
=>×=256
Muy complejo el procedimiento
x = 0 😎
X = 256
How do I solve this: 2^x + 2^(x+1)? Why is it 3* 2^x ? I'm blind, I just can't see it? =D
How do I factor 2^x + 2^(x+1) ?
А если x
ㅉㅊ0/글쿤 ㆍ 3 ⚂ ⛯ ⚅ 2⃣
256
there is a faster and more accurate way:
x > 0, obvious
x cannot be non integer rational number because then LHS = irrational and RHS = rational.
So, x > 1 and x is a natural number.
Let x be 2 to the power of k. k > 0.
So, 2 to the k = 32k.
So, 2 to the (k-5) = k.
Obvious solution: k = 8, so x = 256.
Draw graphs for y = k and y = 2 to the (k-5). For k>8 they cannot meet again (show with differentiation). For 1
@@KyriZee > x > 0, obvious
Why is it obvious? Both 2ˣ and x³² are defined for all x ∈ ℝ.
> x cannot be non integer rational number
But it can be irrational.
> I have no idea how to solve for irrational.
Using the Lambert W function, obviously.
But the solution for an integer x is indeed accurate and neat.
Я применил функцию Ламберта
W(ze^z)=z
2^x=x^32
Извлекаем корень из 32
x=+/- 2^(x/32)
a=e^(Lna)
x=+/-e^[Ln(2^(x/32))]=+/-e^[(xLn2)/32]
xe^[(-xLn2)/32]=+/-1
Домножим обе части на -(Ln2)/32
[(-xLn2)/32]e^[(-xLn2)/32]=+/-(Ln2)/32
Возьмём функцию Ламберта от обеих частей
W{[(-x(Ln2)/32]e^[(-xLn2)/32]}=
W[+/-(Ln2)/32]
-x(Ln2)/32=W[+/-(Ln2)/32]
x=-32W[+/-(Ln2)/32]/Ln2
При помощи калькулятора для функции Ламберта находим
W[+(Ln2)/32]=0,02120634
Ограничимся точностью в 7 значащих цифр
x1=-(0,02120634 Ln2)/32=/0,9790196
W[-(Ln2)/32]=-0,02214590
x2=-(-0,02214590 Ln2)/32=1,02239294
Второе значение
W[-(Ln2)/32]=-5,545177
x3=-(-5,545177 Ln2)/32=256
Последний корень можно получить точно.
Преобразуем выражение -(Ln2)/32 к виду ze^z
-(Ln2)/32=-(2^3Ln2)/2^3*32=-(8Ln2)/2^8=-(8Ln2)/e^(8Ln2)=-(8Ln2)*e^(-8Ln2)
W[-(8Ln2)*e^(-8Ln2)]=-8Ln2
x3=-(-8Ln2)*32/Ln2=256
Bring log...
But at the end of your working; it doesn't make sense ;how took it( 256 the power of 1over 256 is equal x 1 over x ) how you equated
But i appreciate you wow....
Can you say sth on that , what i said