There is also a direct way to find a different solution :take ln to get 32 ln n = n ln2 .Set n= e^t to find -t e^(-t) = - ln( 2/32.) Hence n= exp[-W[-ln 2/ 32]] = 1.02239,. Here W= Lambert W -function.Of course ,your solution looks nicer.
10 print "gt academix-nice olympiad exponential made easy" 20 for a=2 to 10:n=2^a:f=2^n-n^32:if f=0 then 40 30 next a 40 print n:print "2^";n;"=";n;"^32" gt academix-nice olympiad exponential made easy 256 2^256=256^32 > run in bbc basic sdl and hit ctrl tab to copy from the results window.
If we follow the usual use of letters and assume that n is an integer, the answer is only n=256, but if n is a real number, then n≈1.0223929402 or n≈-0.979016935 is also an answer. In that case. x^32 = 2^x is the appropriate expression.
log (n^32) = log (2^n)
32 log n = n log 2
(log n)/n = (log 2)/32
(log n)/n = 8 (log 2)/(32 * 8)
(log n)/n = (log 2^8)/256
(log n) / n = (log 256)/256
So n = 256
@@humbertorodriguezperez1214 Excellent
n=256 is not the only solution.
There are two more!
These are transcendental, and solutions to the equation x^32 = 2^x
Desmos shows two more real solutions: -0.97902 and 1.02239 in addition to 256.
@@padraiggluck2980 no one solution
Use the Lambert W function W(■*e^■) = ■
n^32 = 2^n
ln(n^32) = ln(2^n)
32*ln|n| = n*ln(2) ===> two cases
1st case: n > 0
32*ln(n) = n*ln(2)
ln(n)*n^(-1) = ln(2)/32
ln(n)*(e^ln(n))^(-1) = ln(2)/32
ln(n)*e^(-ln(n)) = ln(2)/32
-ln(n)*e^(-ln(n)) = -ln(2)/32
W(-ln(n)*e^(-ln(n))) = W(-ln(2)/32)
-ln(n) = W(-ln(2)/32)
ln(n) = -W(-ln(2)/32)
n = e^(-W(-ln(2)/32)) ===> -1/e < -ln(2)/32 < 0 ===> 2 real solutions
n₁ = e^(-W₀(-ln(2)/32)) = 1.0223929402057803206527516798494005683768365119132864517728278977...
in WolframAlpha: e^(-productlog(0,-ln(2)/32))
n₂ = e^(-W₋₁(-ln(2)/32)) = 256
in WolframAlpha: e^(-productlog(-1,-ln(2)/32))
2nd case: n < 0
32*ln(-n) = n*ln(2)
ln(-n)*n^(-1) = ln(2)/32
-ln(-n)*n^(-1) = -ln(2)/32
ln(-n)*(-n)^(-1) = -ln(2)/32
ln(-n)*(e^ln(-n))^(-1) = -ln(2)/32
ln(-n)*e^(-ln(-n)) = -ln(2)/32
-ln(-n)*e^(-ln(-n)) = ln(2)/32
W(-ln(-n)*e^(-ln(-n))) = W(ln(2)/32)
-ln(-n) = W(ln(2)/32)
ln(-n) = -W(ln(2)/32)
-n = e^(-W(ln(2)/32))
n = -e^(-W(ln(2)/32)) ===> ln(2)/32 > 0 ===> 1 real solution
n₃ = -e^(-W₀(ln(2)/32)) = -0.979016934957784612322582550011650068748090048886011676265377083...
in WolframAlpha: -e^(-productlog(0,ln(2)/32))
Let n =2^m.
2^(32m) = 2^(2^m)
32m=2^m
(2^5)m=2^m
m = 2^(m-5)
m = 8 = 2^(8-5) = 2^3
n = 2^8 = 256.
😊
@@karelknightmare6712 😊
There is also a direct way to find a different solution :take ln to get 32 ln n = n ln2 .Set n= e^t to find -t e^(-t) = - ln( 2/32.)
Hence n= exp[-W[-ln 2/ 32]] = 1.02239,. Here W= Lambert W -function.Of course ,your solution looks nicer.
@@renesperb Excellent
n^32 = 2^n = (2^x)^(n/x)
2^x = n, n/x = 32
2^8 = 256, 256/8 = 32
n = 256
@@cyruschang1904 Excellent
@GTAcademix Thank you 🙂
n^16^16 n^^2^3^2^3 n^1^1^2^1 n^2^1 (n ➖ 2n+1). 4^2^2 (n ➖ 2n+2).
10 print "gt academix-nice olympiad exponential made easy"
20 for a=2 to 10:n=2^a:f=2^n-n^32:if f=0 then 40
30 next a
40 print n:print "2^";n;"=";n;"^32"
gt academix-nice olympiad exponential made easy
256
2^256=256^32
>
run in bbc basic sdl and hit ctrl tab to copy from the results window.
Allo🎉 n est une puissance de 2 n=2^lp implique 2^n=2^32p implique p=2^(p_5) qui a pour solution p=8 qui me donne n=2^8=256🎉
n^32=(mod12)=8,8×32=256😊
@@sametsahin2130 🤔
Ool
@@sdaqatkahnsdaqatkhan6476 😊
If we follow the usual use of letters and assume that n is an integer, the answer is only n=256, but if n is a real number, then n≈1.0223929402 or n≈-0.979016935 is also an answer.
In that case.
x^32 = 2^x
is the appropriate expression.
@@toshimakusugamo Excellent
Brilliant
@@gabrieldestiny4201 thank You
I don't know how I did it orally 🤓
👨🏫
Teach me this talent plz 🙇♂️🙇♂️
@-ARYABANERJEE give jee 💀
ok, I'm 2/3 genius :)
@@b213videoz oky
@GTAcademix yeah, that was a revalation to me, many thanks to your video
To get a solution (in the set N of numbers), there was a shorter path, it seems to me...
x^32 = 2^x
(x^32)^(1/32x) = (2^x)^(1/32x)
x^(1/x) = 2^(1/32)
recall:
• (a^b)^c = a^(bc) and reciprocally a^(bc) = (a^b)^c
• => 2^(1/32) = (2^8)^(1/32/8)
x^(1/x) = (2^8)^(1/32/8)
x^(1/x) = 256^(1/256)
■ x = 256
🙂
@@GillesF31 Excellent
0
@@TheBorysekPL you are not correct. Try it agsin
@GTAcademix I don't give a damn shit anymore
With such a slow music, very boring
@@MatematikDünyasıBekirHoca so what's your recommend
@ you may talk, explain and solve faster
@MatematikDünyasıBekirHoca okay
Copy frome jj math class!
Sheet
@@Hamza365i i don't know jj🙄
I found 1.02239294
You are not correct.Try it again
@GTAcademix Maple found this solution as well.