assume x= 2 to the power m , so 2^2^m=2^m^32 hence 2^m=m*32 if we put m=8 then 2^8=32*8 which is 256 or 2^2^n=n+5 hence n=3 by considering m=2^n . hence x=2^m=2^8=256
That's how I made it... much faster and smarter than the video ! Notice that there is a negative solution, which is X ~ - 0,979..., easy to prove by studying the two functions 2^x and x^32, but I didn't find the "exact" value... 🙂
Because x is raised to an even power, this equation potentially has three real roots. 2^{-1}0^{32}, so a real root exists between -1 and 0. 2^1>1^{32} and 2^2
Input 2^256 = 256^32 Result True Left hand side 2^256 = 115792089237316195423570985008687907853269984665640564039457584007913129639936 Right hand side 256^32 = 115792089237316195423570985008687907853269984665640564039457584007913129639936 Logarithmic form 256 log_2(2) = 32 log_2(256)
At 7:10 - 7:30 you claim that the unique root of the equation is x=256. This is not correct. See why: The function f(x) = x^(1/x) is not 1-1 and from a relation of the form x^(1/ x)=b^(1/ b) we cannot always conclude x=b The equation may have another root. Eg the equation x^(1/x) =2^(1/2) is true for both x=2 and x=4 , i.e. 2^(1/2)=4^(1/4) but it is not 2=4 Certainly x=b is an obvious root, but this is not a solution of the equation. To solve the equation, all its roots must be found, and in this particular case it was not proved that there are no other roots. Verification is also not necessary. It is certain that the root found from the solution of the equation will verify it. The equation was not completely solved. There are two more roots 1.02239 and -0.97902 It is easy to prove the existence of these roots by applying the Bolzano theorem on the intervals [1, 2] and [-1, 0] for the function f(x)=2^x-x^32 These roots can be found using Lambert functions. For more details and the complete solution of the equation x^(1/x)=a watch my channel named L+M=N at: www.youtube.com/@Nikos_Iosifidis/videos The solution of the equation x^(1/x)=a is at: ua-cam.com/video/oSoJ5IsT_1g/v-deo.html and the video’s name is: Maths for everyone No 46. Study and solution of the equation x^(x^(-1))=a A very similar equation to the equation 2^x=x^32 is solved οn my channel. It is the equation 2^x=x^2 and it also has 3 roots. You can watch this at: ua-cam.com/video/aB0FeExPUsk/v-deo.html I have also presented 2 videos about Lambert functions W_-1 and W_0 One video is called Maths for everyone No 47 - Lambert functions W_0 and W_-1 and located at: ua-cam.com/video/eeW52hA1oc0/v-deo.html This video contains the relevant theory of these functions. The other video is called Maths for everyone No 48 - Solving equations using Lambert functions W_1 and W_0 It is located at: ua-cam.com/video/i5plrWBTX0U/v-deo.html In this video I solve 17 equations using Lambert functions covering almost all cases and explaining the mistakes and omissions in solving these equations. I'd love for you to subscribe to my channel and comment on topics I present.
Exactly right. If you approach this using the Lambert W function it becomes clear that the problem has two solutions! Just because you match the left and right hand sides for a value of x does not mean you can't do the same with other values. The equation x^2 = 2^x works for both x=2 and x=4.
assume x= 2 to the power m , so 2^2^m=2^m^32 hence 2^m=m*32 if we put m=8 then 2^8=32*8 which is 256 or 2^2^n=n+5 hence n=3 by considering m=2^n . hence x=2^m=2^8=256
That's how I made it... much faster and smarter than the video ! Notice that there is a negative solution, which is X ~ - 0,979..., easy to prove by studying the two functions 2^x and x^32, but I didn't find the "exact" value... 🙂
Because x is raised to an even power, this equation potentially has three real roots.
2^{-1}0^{32}, so a real root exists between -1 and 0.
2^1>1^{32} and 2^2
Wowwww amazing 👏 😍 🙌 ❤️ 💖 ♥️ 👏 😍 🙌 ❤️ 💖 ♥️ 👏 😍 🙌 ❤️ 💖 ♥️ 👏 😍 🙌 ❤️ 💖 Explanation ❤❤❤❤❤
Really really nice nice job❤
Input
2^256 = 256^32
Result
True
Left hand side
2^256 = 115792089237316195423570985008687907853269984665640564039457584007913129639936
Right hand side
256^32 = 115792089237316195423570985008687907853269984665640564039457584007913129639936
Logarithmic form
256 log_2(2) = 32 log_2(256)
❤❤
Nice problem! nice solution! ❤❤
Very nice
Nice
2ˣ = x³²
(2ᵃ)ˣ = x³²ᵃ
x = 2ᵃ = 32a
a2⁻ᵃ = 1/32
-a2⁻ᵃ = -1/32 = -(2)⁻⁵
-(2)⁻⁵ = (-2ᵘ)2⁻ᵘ⁻⁵
2ᵘ = u + 5 => u = 3
-(2)⁻⁵ = -8(2)⁻⁸
-a2⁻ᵃ = -8(2)⁻⁸
-a = -8 => a = 8
x = 2ᵃ => x = 2⁸ = 256
4 2^2 (x ➖ 2x+2) x^2^16 x^2^4^4 x^2^4^4 x^2^2^2^2^2 x^1^1^1^1^2 x^1^2(x ➖ 2x+1).
সেই হইসে
X=1.022
At 7:10 - 7:30 you claim that the unique root of the equation is x=256.
This is not correct. See why:
The function f(x) = x^(1/x) is not 1-1 and from a relation of the form x^(1/ x)=b^(1/ b) we cannot always conclude x=b
The equation may have another root. Eg the equation x^(1/x) =2^(1/2) is true for both x=2 and x=4 , i.e. 2^(1/2)=4^(1/4) but it is not 2=4
Certainly x=b is an obvious root, but this is not a solution of the equation. To solve the equation, all its roots must be found, and in this particular case it was not proved that there are no other roots.
Verification is also not necessary. It is certain that the root found from the solution of the equation will verify it.
The equation was not completely solved. There are two more roots 1.02239 and -0.97902
It is easy to prove the existence of these roots by applying the Bolzano theorem on the intervals [1, 2] and [-1, 0] for the function f(x)=2^x-x^32
These roots can be found using Lambert functions.
For more details and the complete solution of the equation x^(1/x)=a watch my channel named L+M=N at: www.youtube.com/@Nikos_Iosifidis/videos
The solution of the equation x^(1/x)=a is at: ua-cam.com/video/oSoJ5IsT_1g/v-deo.html and the video’s name is: Maths for everyone No 46. Study and solution of the equation x^(x^(-1))=a
A very similar equation to the equation 2^x=x^32 is solved οn my channel. It is the equation 2^x=x^2 and it also has 3 roots. You can watch this at: ua-cam.com/video/aB0FeExPUsk/v-deo.html
I have also presented 2 videos about Lambert functions W_-1 and W_0
One video is called Maths for everyone No 47 - Lambert functions W_0 and W_-1
and located at: ua-cam.com/video/eeW52hA1oc0/v-deo.html
This video contains the relevant theory of these functions.
The other video is called Maths for everyone No 48 - Solving equations using Lambert functions W_1 and W_0
It is located at: ua-cam.com/video/i5plrWBTX0U/v-deo.html
In this video I solve 17 equations using Lambert functions covering almost all cases and explaining the mistakes and omissions in solving these equations.
I'd love for you to subscribe to my channel and comment on topics I present.
X=256 final answer
х=1,022
I solved in half page
r u Castilian? if so i am glad.
2 to the power root 2 is rational number or irrational number please reply
Its trancedental
It is irrational
Есть ещё 2 решения:
X=1,022393
X= --0,979017
.....JUST DIFF/INTEGRAL AND IT'LL BE faster
........I JUST *hate* LINEAR ALGEBRA
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...
ln2/2^5 , /*2^3/2^3 , 2^3*ln2/2^8 , 8*ln2*e^(-)8*ln2=lnx*e^(-lnx) , /*(-1) , -8*ln2*e^(-)8*ln2=-lnx*e^(-lnx) ,
case 1 , -8*ln2=-lnx , ln2^8=lnx , x1=2^8 , x=256 ,
case 2 , W(-ln2/32)=-lnx*e^(-lnx) , W(-ln2/32)=-0.022145899 , -0.022145899=-lnx , x=e^0.022145899 , x2=1.02239 ,
test , x1=256 , 2^256=1.15792*10^77 , x1^32=1.15792*10^77 , OK ,
test , x2=1.02239 , 2^1.02239=2.03129 , 1.02239^32=2.03129 , OK ,
Exactly right. If you approach this using the Lambert W function it becomes clear that the problem has two solutions! Just because
you match the left and right hand sides for a value of x does not mean you can't do the same with other values. The equation x^2 = 2^x
works for both x=2 and x=4.
...ORJUST USE DOUBLE DIFF
.......BASIC EXAMPLE OF 1ST COURSE AT MATH ANALYSIS
I stopped watching as soon as you copied the equation