Solving a Nice Exponential Equation with

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Solving a Nice Exponential Equation with

Рет қаралды 2,143

Ай бұрын

Solving a Nice Exponential Equation with ‪@drpkmath1234‬ | Algebra
Join us for an exciting algebraic video as we tackle a nice exponential equation with @drpkmath1234! This problem will raise your understanding about exponentials and problem-solving skills. Whether you're preparing for a math competition or simply love a good math puzzle, this video is perfect for you.
Pause the video and try solving the problem on your own before watching the solution. Let's crack this exponential equation challenge together!
#mathchallenge #exponentialequations #algebra #matholympiad #problemsolving #learnmaths #education #drpkmath1234 #math
Topics covered:
Exponential equations
How to solve exponential equations?
Algebra
Properties of exponents
Properties of surds
Algebraic identities
Radicals
Logarithms
Exponential Equation
Math Olympiad preparation
Math Olympiad training
Exponent laws
Properties of logarithms
Real solutions
Additional Resources:
• A Nice Exponential Equ...
• A Nice Factorial Expon...
• A Fascinating Radical ...
• Solving a Tricky Expon...
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Пікірлер: 8

@tejpalsingh366 Ай бұрын

Comparing well on both sides give x= 1/2 the soln.

@RajeshKumar-wu7ox Ай бұрын

X=1/2, log(root (3+4√3)-√3)/log3 ,log(root(3-4√3)-√3)/log3

@user-kp2rd5qv8g Ай бұрын

Let 3^x=t. Then, t^3-(3+√3)t+3=0. So t=√3 is a solution > x=1/2. Again, if x is real, t>0. Now, [t^3-(3+√3)t+3]/(t-√3)= t^2 +√3t-√3 and t^2 +√3t-√3=0 > t = 1/2[√(3+4√3) -√3] and hence x = 1/(ln3) ln {1/2[√(3+4√3) -√3] }. So, x = 1/2, 1/(ln3) ln {1/2[√(3+4√3) -√3] }.

@SidneiMV Ай бұрын

it's a cubic equation 3ˣ = u u³ + 3 = u(3 + √3) ..... I don't know how to continue .....

@abcekkdo3749 Ай бұрын

X=0.5

@RealQinnMalloryu4 Ай бұрын

{6^6x^2 +6}= 6^12x^2 3^2^3^4x^2 1^3^2^2x^1 ^3^2^1^2x 3^1^1^2x 3^2x (x ➖ 3x+2).3^2x^2+ {6x^2+1}/2= 7x^2/2 {3^2x^2+7x^2/2} = {10x^4/2}= 5x^2 5^1x^2 1^1x^2 1x^2 (x ➖ 2x+1).

@user-ny6jf9is3t Ай бұрын

Αν y=3^χ >0 τοτε χ=1/2 στο Q ενω στο R επι πλεον και χ=[log(-ριζα3+ριζα(3+4ριζα3))-log2]/log3

@walterwen2975 Ай бұрын

Solving a Nice Exponential Equation: 3³ˣ + 3 = 3ˣ⁺¹ + 3ˣ⁺¹⸍², x ϵR; x = ? 3³ˣ - 3ˣ⁺¹ - 3ˣ⁺¹⸍² + 3 = [3³ˣ - 3(3ˣ)] - [(√3)(3ˣ) - 3] = 0 Let: y = 3ˣ > 0, a = √3; [3³ˣ - 3(3ˣ)] - [(√3)(3ˣ) - 3] = (y³ - a²y) - (ay - a²) = 0 y(y² - a²) - (ay - a²) = y(y - a)(y + a) - a(y - a) = (y - a)(y² + ay - a) = 0 y - a = 0; y = a or y² + ay - a = 0; y = [- a + √(a² + 4a)]/2 > 0 y = 3ˣ = a = √3 = 3¹⸍²; x = 1/2 or y = 3ˣ = [- √3 + √(3 + 4√3)]/2 3ˣ = [- √3 + √(3 + 4√3)]/2 = 1.419/2 = 0.709, x = log₃0.709 = - 0.313 Answer check: 3³ˣ + 3 = (3ˣ)³ + 3, 3ˣ⁺¹ + 3ˣ⁺¹⸍² = 3ˣ(3 + √3) x = 1/2, 3ˣ = √3 (√3)³ + 3 = 3√3 + 3, (√3)(3 + √3) = 3√3 + 3; Confirmed x = - 0.313, 3ˣ = 0.709 (0.709)³ + 3 = 3.356, (0.709)(3 + √3) = 3.355; Confirmed The calculation was achieved on a smartphone with a standard calculator app Final answer: x = 1/2 or x = - 0.313