Can You CRACK This TRICKY Factorial Exponential Equation?

Подписаться 6 тыс.

50% 1

Can You CRACK This TRICKY Factorial Exponential Equation?
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.
🌟 Subscribe for more Olympiad math challenges and solutions!
Topics covered:
Factorial equations
Exponential equation
Factorial
Factorial formula
How to solve exponential equations?
Algebra
Properties of exponents
Algebraic identities
Factorial Exponential Equation
Math Olympiad preparation
Math Olympiad training
Exponent laws
Real solutions
Additional Resources:
• A Fascinating Radical ...
• Solving a Tricky Expon...
• Let's Solve A Challeng...
• Tough Exponential Equa...
#factorial #exponentialequations #olympiadmath #mathematics #math #matholympiad #problemsolving #mathchallenge #radical #algebra
Don't forget to like this video if you found it helpful, subscribe to our channel for more Olympiad-focused content, and ring the bell to stay updated on our latest math-solving sessions.
Thanks for Watching !!

Опубликовано:

10 окт 2024

Ссылка:

Скачать:

Готовим ссылку...

Добавить в:

Мой плейлист

Посмотреть позже

Комментарии : 8

@RashmiRay-c1y День назад

Let f(x)=25-x^2 and g(x)=24^(x^2). f(x) is a parabola, symmetric about the y-axis, whose maximum value is f(0)=25 and it is a monotonically decreasing function of x. g(x) is a monotonically increasing function of x, with a minimum value 0f g(0)=1, symmetric about the y-axis. Thus the equation will have at most two solutions (for equal and opposite values of x). By inspection, x= +/-1.

@aurochrok634 12 часов назад

25-x^2 = 24^(x^2) . set y = x^2 and you get 24^y + y = 25. if y>1 then 24^y > 24 and 24^y + y > 24 + 1 = 25. if y

@RashmiRay-c1y День назад

More formally, we can write the equation as 25-x^2 = (1/24)^(25-x^2-25) > (25-x^2)(24)^(25-x^2)= 24^25 > (25-x^2) e^[(25-x^2)ln24] =24^25 > [(25-x^2)ln24] e^ [(25-x^2)ln24] = 24^25 ln 24 > (25-x^2)ln24 = W(24^25 ln 24) > x^2 = 25 - 1/(ln 24) W(24^25 ln 24) = 1. So, x = +/-1. Here W is the Lambert W function.