A Wonderful Factorial Exponential Equation | Olympiad Prep Challenge

[@Quest3669]

X= 3; -3+ - √5 x cant be -ve
Hence 3 is only soln.

[@RashmiRay-c1y]

The equation may be written as 2^[(x-1)! (16x+12 -x(x+1)(x+2)] = 1. x must be a positive integer. Thus, (x-1)! is not zero > x^3+3x^2-14x-12=o which has only one integral solution x=3. Thus, x=3.

[@walterwen2975]

Olympiad Prep Challenge:
256^x! = √[2^(x + 2)!]/[64^(x - 1)!], x, ϵZ ≥ 1; x = ?
256^x! = 2^(8x!) = √[2^(x + 2)!]/[64^(x - 1)!] = √[2^(x + 2)!]/[2^6(x - 1)!]
√[2^(x + 2)!] = [2^(8x!)][2^6(x - 1)!], 2^[(x + 2)!/2] = 2^[8x! + 6(x - 1)!]
(x + 2)!/2 = 8x! + 6(x - 1)!, (x + 2)! = 16x! + 12(x - 1)!
(x + 2)(x + 1)(x)(x - 1)! = 16x(x - 1)! + 12(x - 1)! = (16x + 12)(x - 1)!; x ≥ 1
(x + 2)(x + 1)(x) = 16x + 12, x(x² + 3x + 2) = x³ + 3x² + 2x = 16x + 12
x³ + 3x² - 14x - 12 = x(x² + 3x - 18) + 4(x - 3) = x(x - 3)(x + 6) + 4(x - 3) = 0 (x - 3)(x² + 6x + 4) = 0, x, ϵZ ≥ 1, x² + 6x + 4 > 0; x - 3 = 0, x = 3
Answer check:
256^x! = 2^(8x!) = 2^[(8)(6)] = 2^48
√[2^(x + 2)!]/[64^(x - 1)!]√[2^(3 + 2)!]/[64^(3 - 1)!] = √(2^5!)/(64^2)
= √(2^120)/(2^12) = 2^(60 - 12) = 2^48; Confirmed
Final answer:
x = 3

[@kassuskassus6263]

We have to solve for x the equation x^3+3x^2-14x-12=0 obtained after simplification of the original (cross multiply and raise both sides to the second power). The only solution is x=3. Others are rejected.

[@gnanadesikansenthilnathan6750]

We can go with hit and trial method after framing (x+2) fact = 16x fact + 12(x-1) fact .
This is how I verified the answer and I tried getting the answer.

[@mohammedsaysrashid3587]

A wonderful introduction and clearly explaining thank you Sir🙏 for sharing....x=3

[@RajeshKumar-wu7ox]

3,-3+√5,-3-√5

[@abcekkdo3749]

X=3

[@Fjfurufjdfjd]

Χ=3

[@RealQinnMalloryu4]

8/64=8 2^3 2^1;(x ➖ 2x+1) 25^25 5^5 5^5 1^1 5^1 5^1 (x ➖ 5x+1)