Hungary | A Nice Exponential Algebra Problem | Math Olympiad

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[@강정모-y3l]

Thanks for solving a difficult problem😊

[@عبدالواسع-س8م]

Thank you so much for solving such a challenge problem!

[@ArwindSah]

Easy way to solve this equation (8^x + 27^×)/(12^x +18^x) = 7/6 Simplify 8^x =(2^3)^x =(2^x)^3 27^x = (3^3)^3 = (3^x)^3 12^x =((4)(3))^x =(4^x)(3^x) =((2^2)^x)(3^x) =((2^x)^2)(3^x) Similarly 18^x = ((2)(9))^x =(2^x)(9^x) = (2^x)((3^2)^x) = (2^x)((3^x)^2 Let 2^x =a and 3^x = b then a/b =(2^x)(3^x) a/b = (2/3)^x. Eq. 2 And given equation will look like (a³ + b³)/(a²b +ab²)=7/6 (a+b)(a²-ab+b2)/((ab(a+b)) =7/6 Or (a² - ab +b²)/ab = 7/6 Make two equation from this a² - ab +b² = 7. Eq3 ab = 6 eq.4 Eq. 3 Can be written as a²-ab+b² = 7 (a+b)² - 3ab =7 (a+b)² =7 +3ab = 7+3(6) = 7+18 =25 a+b = √25 = ± 5 Since a =2^x and b =3^x should be positive negative value for a+b is rejected. Thus a + b = 5. Eq. 5 (a-b)² = (a+b)² - 4ab = 5² - 4(6) =25 -24 = 1 a - b = √ 1 = ± 1 eq 6 Case 1. Case 2 a+b = 5. a + b =5 a - b = 1. a - b = - 1 Solving these equation we get (a,b)= (3,2). (2,3) Calculate a/b = 3/2 2/3 Recall value of a b for a/b Case 1 (2/3)^x =3/2 =(2/3)^(-1) Hence x = -1 Case 2 (2/3)^x = 2/3 = (2/3)¹ Hence x =1 Thanks

[@yurenchu]

(8^x + 27^x)/(12^x + 18^x) = 7/6 ... divide numerator and denominator by 8^x ... (1 + (27/8)^x)/((12/8)^x + (18/8)^x) = 7/6 (1 + (27/8)^x)/((3/2)^x + (9/4)^x) = 7/6 (1 + [(3/2)^x]^3 )/((3/2)^x + [(3/2)^x]^2) = 7/6 ... substitute u = (3/2)^x ... (1 + u^3 )/(u + u^2) = 7/6 ... cross-multiply ... 6*(1 + u^3 ) = 7*(u + u^2) 6 + 6u^3 = 7u + 7u^2 ... Upon inspection, u = -1 is a solution. So bring all terms to LHS and factor out (u+1) ... 6 - 7u - 7u^2 + 6u^3 = 0 (1+u) * (6 - 13u + 6u^2) = 0 (u + 1) * (6u^2 - 13u + 6) = 0 (u + 1) * (6u^2 - 4u - 9u + 6) = 0 (u + 1) * (2u(3u - 2) - 3(3u -2)) = 0 (u + 1) * (2u - 3) * (3u - 2) = 0 (u + 1) = 0 OR (2u - 3) = 0 OR (3u - 2) = 0 u = -1 OR u = 3/2 OR u = 2/3 (3/2)^x = -1 OR (3/2)^x = 3/2 OR (3/2)^x = 2/3 x = ln(-1)/ln(3/2) OR x = 1 OR x = -1 x = ln(e^[i*(1+2n)pi])/ln(3/2) , for any integer n OR x = 1 OR x = -1 x = i*pi*(1+2n)/ln(3/2) , for any integer n OR x = 1 OR x = -1 EDIT : The solution from u = -1 is an extraneous solution, leading to 0/0 in the LHS of the original equation. Hence the solution x = i*pi*(1+2n)/ln(3/2) must be dismissed, and the only solutions are x = 1 and x = -1 .

[@Avinashsirkachele43]

Ye 4th hame bata rehe ho

[@erikmoss8972]

Why is hat math's #1

[@wicromaev]

i did it as 35/30=(7^1/x)/(6^1/x), 7/6=(7^1/x)/(6^1/x), x=1 i missed -1