An Interesting Exponential Equation | Olympiad Prep | Algebra

An Interesting Exponential Equation | Olympiad Prep | Algebra
Welcome to infyGyan !! In this video, we'll be solving a nice exponential equation that's both challenging and fun. Perfect for algebra enthusiasts and students preparing for advanced math competitions, this tutorial will take you step-by-step through the process, helping you understand the concepts and techniques needed to tackle exponential equations with confidence.
Join us as we explore this intriguing problem and enhance your algebra skills. If you find the video helpful, don't forget to like, subscribe, and hit the bell icon for more exciting math challenges and tutorials.
Topics covered:
Exponential equations
How to solve exponential equations?
Algebra
Properties of exponents
Properties of surds
Algebraic identities
Radicals
Graphs
Calculus
Lambert W function
Exponential Equation
Math Olympiad preparation
Math Olympiad training
Exponent laws
Real solutions
#educational #exponentialequations #olympiadmath #mathematics #math #matholympiad #problemsolving #radical #algebra #calculus
Additional Resources:
• A Fascinating Radical ...
• Solving a Tricky Expon...
• Let's Solve A Challeng...
• Tough Exponential Equa...
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Thanks for Watching !!

[@RajeshKumar-wu7ox]

√5 is irational number. Sum of two rational numbers is rational.2 is irational number and x^x is irational number. Rational + irational= irrational

[@RyanLewis-Johnson-wq6xs]

2+x^(3x)=Sqrt[5] x = e^W(1/3 (ln(sqrt(5) - 2) + 2 i π n)), i (2 π n - i ln(sqrt(5) - 2))!=0, n element Z

[@gregevgeni1864]

Via the Lambert function :
W( t • e^t) = t (*)
The initial equation equivalent with
x^(3x) = √5 - 2 , x > 0 .
ln (x^(3x)) = ln (√5 - 2)
3 • x • ln x = ln (√5 - 2)
x • ln x = (1/3) • ln (√5 - 2)
x • ln x = ln ([(√5 -2)]^(1/3))
(e^(ln x)) • ln x = ln ([(√5 - 2)]^(1/3)), because x = e^(ln x) (**) .
From (*) W ( (ln x) • e ^(ln x)) = W(( ln[(√5 - 2)]^(1/3))=>
ln x = W((ln [(√5 - 2)] ^(1/3))=>
e^(ln x) = e^{W((ln [(√5 - 2)]^(1/3))}
x = e^{W((ln [(√5 - 2)]^(1/3))} due to (**) .
Of course, we could simplify the cubic root (√5 - 2)^(1/3) ..
Indeed it holds
(√5 - 2)^(1/3) = (√5 - 1)/2 ( exercise :) ).

[@user-kp2rd5qv8g]

Let t=3x. The equation can be written as x^x= (√5-2)^1/3 > xlnx = 1/3 ln(√5-2) > lnx = W(1/3(ln(√5-2)) > x = e^[W(1/3ln(√5-2))], which yields, x= 0.331 +/- 0.3 i, approximately.

[@michaeldoerr5810]

That seems like a much better explanation than the entire video!!!

[@SidneiMV]

x³ˣ = √5 - 2
xˣ = ∛(√5 - 2)
xlnx = (1/3)ln(√5 - 2)
W(xlnx) = W[(1/3)ln(√5 - 2)]
lnx = W[(1/3)ln(√5 - 2)]
*x = e^W[(1/3)ln(√5 - 2)]*

[@RealQinnMalloryu4]

{2+2 ➖ }=4 x^{3+3x ➖ }= {x^6x^2x+x+x+x+x+x x ➖ x ➖ x ➖ x ➖ x ➖ x}=x6x^2 {4+x^6x^2}=4x^6x^2 2^2x^3^2x^2 1^1x^3^1x^2 3x^2 (x ➖ 3x+2)..