An Amazing Factorial Equation Challenge | Can You Solve This?

An Amazing Factorial Equation Challenge | Can You Solve This?

Join us in the algebraic video on an intriguing factorial exponential equation, perfect for Math Olympiad preparation! In this video, we explore the complexities and solutions of this challenging mathematical problem, offering insights and techniques essential for tackling Olympiad-level mathematics.

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Topics covered:

Factorial equations
Exponential equation
Factorial
Factorial formula
How to solve exponential equations?
Algebra
Properties of exponents
Algebraic identities
Radicals
Factorial Exponential Equation
Math Olympiad preparation
Math Olympiad training
Exponent laws
Real solutions

Additional resources:

#factorial #exponentialequations #olympiadmath #mathematics #math #matholympiad #problemsolving #mathchallenge #radical #algebra

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[RashmiRay-cy]

Setting 2x+3=t, the given equation is tantamount to (t-1)t(t+1)+2(t+1)=12 > t^3+t-10 =0. t=2 is the only real solution. Thus, 2x+3=2 > x=-1/2.

[RealQinnMalloryu]

{1x+1x ➖ }/{20x+20x ➖ }+{10+10 ➖ }{3x+3x ➖ 1x^1/6^11x^4 /6^11^1x^4 /6^1^1x^4 /3^2x^2^2 /3^1x1^2 /3x^2 (x ➖ 3x+2). {3x+3x ➖ }/{200x+200x ➖ }+{400+400 ➖ }+{6x+6x ➖ }= 3^2x^2/3^4^3^4x^4 1^1x^1/1^2^23^2^2x^2^2 /1^13^1^1x^1^2 /3x^2 (x ➖ 3x+2). {2x+2x ➖ }/{20x+20x ➖ }+{30+30 ➖ }+{6x+6x ➖ 2^2x^2/10^10^12x^4 1^1x^1/2^5^2^5^3^4x^4/ 1^1^1^1^3^2^2x^2^2 /3^1^1x^1^2 3x^2 (x ➖ 3x+2).

[key_board_x]

Numerator on the right side
100^[3! / (2x + 4)!] = [10^(2)]^[3! / (2x + 4)!]
100^[3! / (2x + 4)!] = 10^[(2 * 3!) / (2x + 4)!]
100^[3! / (2x + 4)!] = 10^[(2 * 3 * 2 * 1) / (2x + 4)!]
100^[3! / (2x + 4)!] = 10^[12 / (2x + 4)!]
100^[3! / (2x + 4)!] = 10^[12 / (2x + 4).(2x + 3).(2x + 2).(2x + 1)!]
100^[3! / (2x + 4)!] = 10^[12 / 2.(x + 2).(2x + 3).(2x + 2).(2x + 1)!]
100^[3! / (2x + 4)!] = 10^[12 / 4.(x + 2).(2x + 3).(x + 1).(2x + 1)!]
100^[3! / (2x + 4)!] = 10^[3 / (x + 2).(2x + 3).(x + 1).(2x + 1)!] → let: a = (2x + 3).(x + 1).(2x + 1)!
100^[3! / (2x + 4)!] = 10^[3/{a.(x + 2)}]
Denominator on the right side
10^[2! / (2x + 3)!] = 10^[2 / (2x + 3)!]
10^[2! / (2x + 3)!] = 10^[2 / (2x + 3).(2x + 2).(2x + 1)!]
10^[2! / (2x + 3)!] = 10^[2 / 2.(2x + 3).(x + 1).(2x + 1)!]
10^[2! / (2x + 3)!] = 10^[1 / (2x + 3).(x + 1).(2x + 1)!] → recall: a = (2x + 3).(x + 1).(2x + 1)!
10^[2! / (2x + 3)!] = 10^(1/a)

Ratio on the right side
100^[3! / (2x + 4)!] / 10^[2! / (2x + 3)!] = 10^[3/{a.(x + 2)}] / 10^(1/a)
100^[3! / (2x + 4)!] / 10^[2! / (2x + 3)!] = 10^{ [3/{a.(x + 2)}] - (1/a) }
100^[3! / (2x + 4)!] / 10^[2! / (2x + 3)!] = 10^{ [3 - (x + 2)] / a.(x + 2) }
100^[3! / (2x + 4)!] / 10^[2! / (2x + 3)!] = 10^{ (1 - x) / a.(x + 2) }

On the left side
10^[1! / (2x + 1)!] = 10^[1/(2x + 1)!] → recall: a = (2x + 3).(x + 1).(2x + 1)! → (2x + 1)! = a/[(2x + 3).(x + 1)]
10^[1! / (2x + 1)!] = 10^[1/{ a/[(2x + 3).(x + 1)] }]
10^[1! / (2x + 1)!] = 10^[(2x + 3).(x + 1)/a]

The original equation
10^[1! / (2x + 1)!] = { 100^[3! / (2x + 4)!] } / { 10^[2! / (2x + 3)!] } → according to the previous results
10^[(2x + 3).(x + 1)/a] = 10^{ (1 - x) / a.(x + 2) }
(2x + 3).(x + 1)/a = (1 - x) / a.(x + 2)
(2x + 3).(x + 1) = (1 - x)/(x + 2)
(2x + 3).(x + 1).(x + 2) = 1 - x
(2x + 3).(x² + 3x + 2) = 1 - x
2x³ + 6x² + 4x + 3x² + 9x + 6 = 1 - x
2x³ + 9x² + 14x + 5 = 0 → the aim is to make terms to the 2nd power disappeared → let: x = z - (3/2)
2.[z - (3/2)]³ + 9.[z - (3/2)]² + 14.[z - (3/2)] + 5 = 0
2.[z - (3/2)]².[z - (3/2)] + 9.[z² - 3z + (9/4)] + 14z - 21 + 5 = 0
2.[z² - 3z + (9/4)].[z - (3/2)] + 9z² - 27z + (81/4) + 14z - 21 + 5 = 0
2.[z³ - (3/2).z² - 3z² + (9/2).z + (9/4).z - (27/8)] + 9z² - 27z + (81/4) + 14z - 21 + 5 = 0
2.[z³ - (9/2).z² + (27/4).z - (27/8)] + 9z² - 13z + (17/4) = 0
2z³ - 9z² + (27/2).z - (27/4) + 9z² - 13z + (17/4) = 0
2z³ + (1/2).z - (5/2) = 0
4z³ + z - 5 = 0 ← we can see that (1) is an obvious root
z = 1 → recall: x = z - (3/2)
x = 1 - (3/2)
→ x = - 1/2