A Nice Type of Exponential Equation

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[@Don-Ensley]

problem
x³ ˣ^¹⁵ = 5
ln both sides.
3x¹⁵ ln x = ln 5
Replace the first x with eˡⁿ ˣ.
3 (eˡⁿ ˣ) ¹⁵ ln x = ln 5
Bring the 15 inside as a multiple in the exponent.
3 (e¹⁵ ˡⁿ ˣ) ln x = ln 5
Move the ln x to the front after the 3.
3 ln x e¹⁵ ˡⁿ ˣ = ln 5
Multiply both sides by 5.
15 ln x e¹⁵ ˡⁿ ˣ = 5 ln 5
= ln 5 eˡⁿ ⁵
We do not need to use Lambert's W function because positionally we have an exact match between 15 ln x on the left and ln 5 on the right.
It must be that
15 ln x = ln 5
ln x = (1/15) ln 5
ln x = ln 5 ⁽¹ᐟ¹⁵⁾
x = 5 ⁽¹ᐟ¹⁵⁾
answer
x = ¹⁵√5

[@Pascal-cy9sc]

x=5^(1/15)(cos(2/15 nπ)+i sin(2/15 nπ))(n=0,1,····,14)
も複素数を含めて、15個の解を持ちます。

[@fhffhff]

4а-2b+5=0 b=2a+2,5 x=-b/2/a a=-0,25,b=2 S=-0,25x³+4x²-11x-4 -0,75x²+8x-11=0 x=2(8±√31)/3>4 x=2(8+√31)/3 S=-9 7/9+22/27√31

[@abdelmoulamsaddaq8240]

Deuxième méthode ❤

[@-wx-78-]

Notice that any solution of x¹⁵=5 is a solution for x**(3x¹⁵)=5, then ¹⁵√5 comes naturally.

[@Pascal-cy9sc]

正にその通りですね✨私が全ての解を列挙しました。

[@giuseppemalaguti435]

t=W(5ln5)/15...x=e^t=5^(1/15)

[@scottleung9587]

I used the second method.

[@trojanleo123]

Used my trusty friend Lambert W and got the answer
x = 5^(1/15)

[@rogerphelps9939]

The Lambert W function comes up in many videos like this but without exception the nuts and bolts of it. how it is evaluated is conveniently omitted. we are just presented with a black box that we can enter numbers into using some sort of app and hey presto the answer is spat out. This really is rather unsatisfactory.

[@rakenzarnsworld2]

x^(1/15)

[@ViaanHossain]

The first viewer and liker.

[@SyberMath]

😍❤️