An Interesting Problem With Reciprocal Exponents

[@Nothingx303]

Bro your methods of math solving is really good. Are you thinking to start any research in math or to solve any unsolved problem?

[@Qermaq]

Other interesting setups: 4^x + 4^(1/x) = 18, 4^x + 4^(1/x) = 3/4.

[@Nothingx303]

Let x= n for my keyboard 😂
So, 4ⁿ = a
=> take natural log on both sides
So n×ln(4) = ln(a)
=> n =log of base a and arg of 4
Take reciprocal
I/n = log of base 4 and arg of a
So a + 4^ log of base 4 and arg of a
=8
So a + a =8
2a =8
a =4
So 4ⁿ =4
So n=1
Thank you...................😊

[@ronjharedrobles2960]

0:39 by visual
x=1
and idk why

[@josephwellinghoff1259]

1 by inspection

[@MattyFein04]

Legit. Took me

[@Quest3669]

4+4= 8 so 1

[@lucien346]

Yes but how to prove its the only solution

[@Quest3669]

@lucien346 l.h.s is constt rhs increasung hence 1 soln

[@kamarinelson]

No method with quadratic algebra?

[@RealQinnMalloryu4]

{4x+4x ➖}+{4+4 ➖ }1/x/={8x^2+8}1/x/=16x^2*1/x 16x^2/x=16x^2 4^4x^2 2^2^2^2x^2 1^1^1^1x^2 1x^2 (x ➖ 2x+1).