A Wonderful Factorial Exponential Equation | Olympiad Prep Challenge

A Wonderful Factorial Exponential Equation | Olympiad Prep Challenge
Let's dive into the world of advanced mathematics with our latest video, "A Wonderful Factorial Exponential Equation | Olympiad Prep Challenge." This video presents a unique and fascinating problem involving factorials and exponential equations, perfect for those preparing for Math Olympiads or anyone who loves a cool math challenge.
Join us as we break down this complex equation step-by-step, offering valuable tips and strategies to tackle similar problems. Whether you're a seasoned math enthusiast or gearing up for a competitive exam, this video is designed to push your problem-solving skills to the next level.
Don't forget to like, share, and subscribe for more intriguing math challenges and tutorials. Let's solve this wonderful equation together!
Topics covered:
Exponential equations
Factorial
Factorial laws
How to solve exponential equations?
Algebra
Properties of exponents
Algebraic identities
Radicals
Synthetic division method
Rational root theorem
Cubic equations
Factorial Exponential Equation
Math Olympiad preparation
Math Olympiad training
Exponent laws
Integer solutions
Timestamps:
0:00 Introduction
0:44 Exponent laws
4:36 Factorial formula
6:44 Cubic equation
8:30 Synthetic division
8:50 Quadratic equation
9:00 Quadratic formula
9:50 Integer solution
10:25 Verification
Additional Resources:
• An Interesting Factori...
• An Engaging Exponentia...
• A Fascinating Radical ...
• Solving a Tricky Expon...
#factorial #exponentialequations #olympiadmath #mathematics #math #matholympiad #problemsolving #mathchallenge #radical #algebra
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Thanks for Watching !!

[@mohammedsaysrashid3587]

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

[@user-kp2rd5qv8g]

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

[@RajeshKumar-wu7ox]

3,-3+√5,-3-√5

[@tejpalsingh366]

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

[@user-ny6jf9is3t]

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)

[@abcekkdo3749]

X=3