A Fascinating Exponential Equation | Math Olympiad | Algebra

A Fascinating Exponential Equation | Math Olympiad | Algebra
Let's explore the world of algebra with a fascinating exponential equation for Math Olympiad enthusiasts. In this video, we break down the steps to solve a challenging exponential equation, providing various approaches with clear explanations and valuable insights. Perfect for students preparing for math competitions or anyone looking to sharpen their algebra skills. Join us and explore the beauty of exponential equations!
Topics covered:
Exponential equations
Graphs
How to solve exponential equations?
Algebra
Properties of exponents
Algebraic identities
Radicals
Logarithms
Monotonic increasing function
Properties of logarithms
Factorial Exponential Equation
Math Olympiad preparation
Math Olympiad training
Exponent laws
Real solutions
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• An Interesting Algebra...
#radicals #exponentialequations #olympiadmath #mathematics #math #matholympiad #problemsolving #mathchallenge #radical #algebra
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[@tunneloflight]

All of these amount to the same underlying method -> observational. starting with 64, recognizing root 64 = 8, and cube root 64 = 4, and that that resolves the problem.

[@kahnhuang]

Are we saying 64^8^4 = (64^64)^64?

[@RealQinnMalloryu4]

x^3+x^3x^3x^3 ➖ x^3 ➖ x^3=x9 x^3^2 (x ➖ 3x+3). (4^16^4^16)4^16 *(4^4^4^4^4)4^4^4 (2^2^2^2^2^2 2^2^2^2^2^2)^2^2^2^2^2^2 (1^1^1^11^1 1^11^11^1)1^1^1^1^1^2 1^2 (x ➖ 2x+1) .

[@SidneiMV]

x = u⁶
u⁶^(u³^u²) = (64⁶⁴)⁶⁴
(u³^u²)6log₂u = 64log₂64⁶⁴ = 64²6
(u³^u²)log₂u = 64² = 2¹²
u = 2ᵛ => v = log₂u
v(2³ᵛ)^(2²ᵛ) = 2¹² = (2³)^(2²)
v = 1 (by inspection)
log₂u = 1 => u = 2 => *x = 2⁶ = 64*
64^8⁴ = (64⁶⁴)⁶⁴
2^(6¹8⁴) = 64^64² = 2^(6¹64²)
6¹8⁴ = 6¹64² => 8⁴ = 64²
2¹² = (2⁶)² => 2¹² = 2¹² (true)