A Nice Exponential Equation Challenge | An Algebra Puzzle!
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Ай бұрынA Nice Exponential Equation Challenge | An Algebra Puzzle!
Welcome to another exciting algebra challenge! In this video, we'll solve a fascinating exponential equation that will raise your mathematical skills. Can you solve it? Watch the video, try it yourself, and share your solutions in the comments! This is a perfect problem for those preparing for math competitions or anyone who loves a good brain teaser. Don't forget to like, subscribe, and hit the notification bell for more challenging math puzzles!
Topics Covered:
Exponential equations
How to solve exponential equations?
Algebra
Properties of exponents
Algebraic identities
Exponential Equation
Math Olympiad preparation
Math Olympiad training
Exponent laws
Solving quintic equation
Quadratic equation
Discriminant
Real solutions
Additional resources:
• An Interesting Algebra...
• A Nice Algebra Problem...
• Can You Solve This Exp...
• Solving a Tricky Expon...
#matholympiad #exponentialequations #problemsolving #mathcompetition #mathtutorial #mathematics #learnmaths #mathenthusiast #algebra #exponents #algebrachallenge
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Thanks for Watching !!
@ZhilinChen-my7tp 29 күн бұрын
X=0, (log3+√5/ log 2)-1, (log3-√5/ log 2)-1
@RajeshKumar-wu7ox 29 күн бұрын
X=0,log((3+√5)/2)/log2
@tejpalsingh366 29 күн бұрын
x=0; (log(3+-√5)/ log2)-1
@kassuskassus6263 29 күн бұрын
Let u=2^x and solve for u. u=1 (double) gives x=0 and u=(3+or-sqrt5)/2 gives x=log((3+or-sqrt5)/log2)-1.
@user-kp2rd5qv8g 29 күн бұрын
Let t=2^x. Then, [t^4+8t^2+1]/[[t^3+t] = 5 > t^2[t^2+1/t^2+8]/[t^3+t] = 5 > [(t+1/t)^2+6]/(t+1/t) = 5 > (t+1/t)^2 -5 (t+1/t) + 6 = 0 > t+1/t = 2,3 > x=0 or x = log_2 [(3 +/-sqrt5)/2].
@prateek1.9 28 күн бұрын
cant you use hindi sir??
@user-ny6jf9is3t 29 күн бұрын
χ=0 διπλη, ή χ=[[log(3+ -ριζα5)]/log2]-1
@SidneiMV 28 күн бұрын
2^x = u (u⁴ + 8u² + 1)/(u³ + u) = 5 [(u² + 1)² + 6u²]/(u³ + u) = 5 (u² + 1)/u + 6u/(u² + 1) = 5 (u² + 1)/u = w w + 6/w = 5 w² - 5w + 6 = 0 (w - 2)(w - 3) = 0 w = 2 (u² + 1)/u = 2 u² - 2u + 1 = 0 (u - 1)² = 0 => u = 1 2^x = 1 => *x = 0* w = 3 (u² + 1)/u = 3 u² - 3u + 1 = 0 u = (3 ± √5)/2 2^x = (3 ± √5)/2 2^(x + 1) = 3 ± √5 (x + 1)ln2 = ln(3 ± √5) *x = (1/ln2)ln(3 ± √5) - 1*
@RealQinnMalloryu4 28 күн бұрын
{4^4x^24^6x^2}={ 8^10x^4+2}= 10^10x^4=/{4^6x^2+4x^2}= 8^6x^4 _10^10x^4/8^6^x^4=1.^2^1^.2^6x^1 1^1.2^1^1^1.2^1^3^2x1^1 11 x3^2 x^3^2.(x ➖ 3x+2 )
@user-nd7th3hy4l 28 күн бұрын
X=0
@abcekkdo3749 29 күн бұрын
X=0,
@yakupbuyankara5903 29 күн бұрын
X=0
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