Nice Exponential Problem | Olympiad Math Algebra Question

[@prabhudasmandal6429]

Good logical solution.

[@MathBeast.channel-l9i]

Thanks 😊

[@AljosaVolcic]

As noted, you did not find all the solutions. So what does your "verification" tell us?

[@MathBeast.channel-l9i]

It tell us verification about the solution we have gotten in this tutorial 🫣

[@DedMatveev]

There are not 32 roots,only 3😅. 2 not found.

[@payoo_2674]

a^32 = 2^a
ln(a^32) = ln(2^a)
32*ln|a| = a*ln(2) ===> two cases
1st case: a > 0
32*ln(a) = a*ln(2)
ln(a)*a^(-1) = ln(2)/32
ln(a)*(e^ln(a))^(-1) = ln(2)/32
ln(a)*e^(-ln(a)) = ln(2)/32
-ln(a)*e^(-ln(a)) = -ln(2)/32
W(-ln(a)*e^(-ln(a))) = W(-ln(2)/32)
-ln(a) = W(-ln(2)/32)
ln(a) = -W(-ln(2)/32)
a = e^(-W(-ln(2)/32)) ===> -1/e < -ln(2)/32 < 0 ===> 2 real solutions
a₁ = e^(-W₀(-ln(2)/32)) = 1.0223929402057803206527516798494005683768365119132864517728278977...
a₂ = e^(-W₋₁(-ln(2)/32)) = 256
2nd case: a < 0
32*ln(-a) = a*ln(2)
ln(-a)*a^(-1) = ln(2)/32
-ln(-a)*a^(-1) = -ln(2)/32
ln(-a)*(-a)^(-1) = -ln(2)/32
ln(-a)*(e^ln(-a))^(-1) = -ln(2)/32
ln(-a)*e^(-ln(-a)) = -ln(2)/32
-ln(-a)*e^(-ln(-a)) = ln(2)/32
W(-ln(-a)*e^(-ln(-a))) = W(ln(2)/32)
-ln(-a) = W(ln(2)/32)
ln(-a) = -W(ln(2)/32)
-a = e^(-W(ln(2)/32))
a = -e^(-W(ln(2)/32)) ===> ln(2)/32 > 0 ===> 1 real solution
a₃ = -e^(-W₀(ln(2)/32)) = -0.979016934957784612322582550011650068748090048886011676265377083...

[@b213videoz]

You can keep your imaginary to yourself 😂

[@richardlongman5602]

If we allow imaginary numbers then there are a countable infinity of roots.

[@quadibloc2]

I thought about the problem by asking what would happen if a was 2, or 32. Those didn't work. Then I realized that 2^a is bigger asymptotically than a^32, so the right answer has to be big. 1024 was too big. And then I realized: 256 equals 2^8, and it also equals 8*32. So 256^32 equals 2^(8*32), which makes it the answer.

[@thewolfdoctor761]

Yes, trial and error works. I took log base 2 of both sides and got 32log(base 2) a = a, then came up with 256 quickly.

[@MathBeast.channel-l9i]

Yes.
Nice Effort by you Sir 🙂

[@ravikiranpalaparthi615]

a³² = 2^a
=> a^{2⁵} = 2^a
a^{2⁵/a} = 2
Let a = 2^x
2^{x*2⁵/a} = 2¹
=> x*2⁵/a = 1
=> x*2⁵/(2^x) = 1
=> x*2⁵ = 2^x
Let x = 8
=> 8*2⁵ = 2⁸
2³*2⁵ = 2⁸ -> x= 8 holds true
So, a = 2^x = 2^8 = 256
Check:
a³² = 2^a
256³² = 2^{256}
2^{8*32} = 2^{256}
2²⁵⁶ = 2²⁵⁶
Hence proved

[@MathBeast.channel-l9i]

Nice Effort Sir
Thanks for sharing 😊

[@ΒασιληςΑρετακης-μ2ψ]

Логарифмируем обе части по основанию 2.получаем 32 логарифм а по основанию 2 равняется а.Далее,обозначаем логарифм а по основанию 2 через t и получаем показательное уравнение 32t=2^t ,решение которого легко подобрать.Оно выполняется при t=8.Графиком t=8.Графиком левой части является прямая y=32 t,правая часть показательная функция y=2^t.Эти графики имеют единственную точку пересечения:t=8.

[@віталійкуценко-л6ц]

2 пересечения же - первое при х чуть больше 1/32 и у чуть больше 1, второе в точке (8;256). К любой экспоненте имеется касательная, проходящая через центр координат, остальные прямые, проходящие через центр и лежащие в положительном квадрате, либо пересекают её в 2 точках, либо не пересекают вообще.

[@joseeoliviero6078]

Nice! For a while, I thought you were going to find all 32 roots. That might take some time lol

[@MathBeast.channel-l9i]

Eventually🫡🙂

[@prollysine]

32*lna=a*ln2 , lna/a=ln2/2^5 , lna*e^(-lna)=2^3*ln2/2^8 , /*(-1) , W(-lna*e^(-lna))=W(-8*ln2*e^(-8*ln2)) ,
-lna=-8*ln2 , lna=ln256 , a1=256 , test , a1^32=~ 1.15792*10^155 , 2^a1=~ 1.15792*10^155 , OK ,
W(-ln2/32)=--lna , a=e^(-W(-ln2/32)) , a2=~ 1.02239 , test , a2^32=~ 2.03129 , 2^a2=~ 2.03129 , OK ,

[@arturdietrich]

like i do, by deduction and inspection !

[@Italianor01]

A=2 a=32. ❤❤

[@piolino1535]

With lambert function ,,

[@Maria-wn4du]

Multumesc ! 👏

[@MathBeast.channel-l9i]

You are welcome 🤗

[@rajkotalovic9784]

Ja ovo nista ne razumem, jeli ovo 1 zadatak?

[@olayaher1]

La resolví por W Lambert, finalmente me quedó una expresión así: W(-Ln(a)*e^(-Ln(a))=W(-Ln(2)/32)), dando como resultado -Ln(a)=-0.0221458994977...., de donde a=e^0.0221458994977...., a=1.02239, y cumple con la igualdad dada

[@payoo_2674]

a^32 = 2^a
ln(a^32) = ln(2^a)
32*ln|a| = a*ln(2) ===> two cases
1st case: a > 0
32*ln(a) = a*ln(2)
ln(a)*a^(-1) = ln(2)/32
ln(a)*(e^ln(a))^(-1) = ln(2)/32
ln(a)*e^(-ln(a)) = ln(2)/32
-ln(a)*e^(-ln(a)) = -ln(2)/32
W(-ln(a)*e^(-ln(a))) = W(-ln(2)/32)
-ln(a) = W(-ln(2)/32)
ln(a) = -W(-ln(2)/32)
a = e^(-W(-ln(2)/32)) ===> -1/e < -ln(2)/32 < 0 ===> 2 real solutions
a₁ = e^(-W₀(-ln(2)/32)) = 1.0223929402057803206527516798494005683768365119132864517728278977...
a₂ = e^(-W₋₁(-ln(2)/32)) = 256
2nd case: a < 0
32*ln(-a) = a*ln(2)
ln(-a)*a^(-1) = ln(2)/32
-ln(-a)*a^(-1) = -ln(2)/32
ln(-a)*(-a)^(-1) = -ln(2)/32
ln(-a)*(e^ln(-a))^(-1) = -ln(2)/32
ln(-a)*e^(-ln(-a)) = -ln(2)/32
-ln(-a)*e^(-ln(-a)) = ln(2)/32
W(-ln(-a)*e^(-ln(-a))) = W(ln(2)/32)
-ln(-a) = W(ln(2)/32)
ln(-a) = -W(ln(2)/32)
-a = e^(-W(ln(2)/32))
a = -e^(-W(ln(2)/32)) ===> ln(2)/32 > 0 ===> 1 real solution
a₃ = -e^(-W₀(ln(2)/32)) = -0.979016934957784612322582550011650068748090048886011676265377083...

[@MathBeast.channel-l9i]

Brilliant effort Boss 🫡

[@RealQinnMalloryu4]

a^2^5 a^2^2^3 a^2^1^1 a^2^1 (a ➖ 2a+1). 2^(5)=32 2^5 2^2^3 2^1^1 2^1 (a ➖ 2a+1).

[@اسماعیلخسروی-خ6ظ]

❤❤❤❤❤

[@MathBeast.channel-l9i]

Thanks ❤️

[@dongweilin5393]

if a≺0, (a^32)^(1/a)=?=(a^(1/a))^(32)

[@sunnysharma5166]

256

[@shang_psycho7414]

This was so satisfying

[@MathBeast.channel-l9i]

Thanks 😊

[@shang_psycho7414]

@ yw

[@ManojkantSamal]

a=256, Explain later

[@MathBeast.channel-l9i]

Okay 👍

[@BajoJajoBajoJajo2]

isn’t this just guess and check which you could do in your head just by looking at the original question?

[@jamespat7975]

But I get, a = 1.0224, so what's wrong ?

[@truonghoang8926]

This method give one solution only, not all solution can be found.

[@віталійкуценко-л6ц]

All is right. If the equation is x^1/x=a, a>1, it has two positive solutions except of a case when a=e^1/e. One solution is less then e, other is more. The classical example is x^1/x=√2, x1=2, x2=4.