Math Challenge l A Nice Exponential Problem l Find the value of X

[@prabhushettysangame6601]

Excellent example 👏

[@RealQinnMalloryu4]

2022(2)^2 ➖ (x)^2 +(2022^{2+2 ➖ }+{x+x ➖}= 2022^{4 ➖ x^2}+2022^{4+x^2}=2022^{x^0+x^0 ➖ x^0 +x^0 ➖}+2022^{4+x^2}=2022^{x^1+x^1}+2022^4x^2=2022^x^2+8088x^2={4044+8088x^2}=12.132x^2 1^1.1^1^1^x^2 1x^2 (x ➖ 2x+1). (2022)^2=40484 2^2^02^2^2^3^2^2 1^10.1^1^1^1^1^1^1 1^2^5 1^2^1 2^1(x ➖ 2x+1).

[@ganeshdas3174]

Go for 2022^2(x +1/x) = 2.2022^2
( x +1/x) = 2
(√x) ^2 +(1/√x) ^2 - 2 x. 1/x = 0
(√x - 1/√x) ^2 = 0
x = 1

[@adribber]

2022²-ⁿ + 2022²+ⁿ = 2(2022)²
2022²(2022-ⁿ + 2022+ⁿ) = 2(2022)²
(2022-ⁿ + 2022+ⁿ) = 2
2022²ⁿ - 2*2022ⁿ + 1 = 0
(2022ⁿ - 1)² = 0
2022ⁿ = 1
∴ n = 0

[@marcgriselhubert3915]

First we simplify by 2022^2, it gives: 2022(-x) + 2022^x = 2 or 2.cosh(x.ln(2022)) = 2, or cosh(x.ln(2022) = 1.
That gives an unique solution: x.ln(2022) = 0 or x = 0. (As cosh(X) = 1 is equivalent to X = 0)