A Beautiful Exponential Equation | Math Olympiad Algebra

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A Beautiful Exponential Equation | Math Olympiad Algebra
Welcome to our Math Olympiad series! In this video, we dive into a beautiful exponential equation that highlights the elegance and complexity of algebra. We'll break down the problem step-by-step, exploring various techniques and strategies to solve it. Whether you're preparing for a math competition or just love a good mathematical challenge, this video is for you. Join us as we uncover the secrets of this exquisite exponential equation and enhance your algebraic skills.
Topics covered:
Exponential equations
How to solve exponential equations?
Algebra
Logarithms
Properties of logarithms
Properties of exponents
Algebraic identities
Exponential Equation
Math Olympiad preparation
Radicals
Exponential Equations with different bases
Math Olympiad training
Exponent laws
Math tutorial
Math Olympiad
Additional resources:
• Solving a Tricky Expon...
• Let's Solve A Challeng...
• Tough Exponential Equa...
• Cracking a Fascinating...
#matholympiad #exponentialequations #problemsolving #math #mathtutorial #mathematics #learnmaths #mathenthusiast #algebra #radical
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Опубликовано:

16 окт 2024

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Комментарии : 7

@mohammedsaysrashid3587 4 месяца назад

It was a wonderful introduction and clearly explaining...thanks for sharing...x=0 ,x= 1.7 06

@kassuskassus6263 4 месяца назад

Let a=2^x and b=3^x and raise both sides to the third power. After simplifications, we'll get (a-b)(2a-b)=0, so a=b (means that 2^x=3^x, so x=0) and 2a=b, so 2*2^x=3^x. Log both sides, we get x=log2/log3/2)

@RashmiRay-c1y 4 месяца назад

The given expression can be written as (2^x-3^x)^1/3 [(2^x-3^x)^2/3 - (2^x)^2/3] =0. If the first factor is zero, 2^x=3^x > x=0. The second factor is [(2^x -3^x)^1/3]^2 -[(2^x)^1/3]^2=0, which can be written as [(2^x-3^x)^1/3 + 2^x/3][(2^x-3^x)^1/3 - 2^x/3] =0. if the first factor is zero > 2^(x+1) = 3^x > x= ln 2/(ln 3 - ln 2). The second factor does not vanish for any finite x. Thus, x = 0, ln 2/(ln 3 - ln 2).

@Fjfurufjdfjd 4 месяца назад

χ=0 ή χ=log2/(log3-log2)

@tunneloflight 4 месяца назад

If you remove the constraint on the solution being a real number, the third root is negative infinity.

@paulortega5317 2 месяца назад

let u = (2^x - 3^x)/2^x x = log(1 - u)/(log(3) - log(2)) After substitution with u, u^2 = u u= 0, 1, -1 for u = 0 x = 0 for u = -1 x = log(2)/(log(3) - log(2)) for u = 1, no valid x