Poland Math Olympiad l A Very Nice Algebra Problem

[@prollysine]

let u=(17/7)^x , u > 0 , u^2-u-1=0 , u=(1+/-V5)/2 , (17/7)^x=(1+V5)/2 , x=log((1+V5)/2)/log(17/7) ,
test , 49^x+119^x=~ 21.6083 , 289^x=~ 21.6083 , OK ,

[@vijaymaths5483]

Thank you

[@adribber]

49ⁿ + 119ⁿ = 289ⁿ
(7²)ⁿ + (7*17)ⁿ = (17²)ⁿ
(7ⁿ)² - (17ⁿ)² = - (7ⁿ*17ⁿ)
(7ⁿ)²/(7ⁿ*17ⁿ) - (17ⁿ)²/(7ⁿ*17ⁿ) = - 1
7ⁿ/17ⁿ - 17ⁿ/7ⁿ = - 1
(17/7)ⁿ - (7/17)ⁿ = 1
(17/7)ⁿ - 1/(17/7)ⁿ = 1
put, (17/7)ⁿ = k > 0
k - 1/k = 1
k² - k - 1 = 0
k = [- (- 1) ± √{(- 1)² - 4*1*(- 1)}] / 2*1 = (1 ± √5)/2
∴ k = (1 + √5)/2
Recall, (17/7)ⁿ = k
(17/7)ⁿ = (1 + √5)/2
log(17/7)ⁿ = log(1 + √5)/2
nlog(17/7) = log(1 + √5)/2
n = [log(1 + √5)/2] / log(17/7)
∴ n = [log(1 + √5) - log2] / [log17 - log7]