A awesome mathematics exponents problem | Olympiad Question | can you solve this problem | x=? 

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You made a mistake. Log 13 is equal to 1.114. Log 11 is equal to 1.0413. The correct answer is x = 1.8017

The answer is 2.8787. Golden equations ALWAYS give the BEST decimal answers!!!

68 4^17 4^17^1 2^2^1^1 1^2 (x ➖ 2x+1).

The correct answer is 1.793799. There is a mistake with the calculation of log(17)-log(13)

x=2.88 not a solu , 289^(2.88)-221^(2.88)=6.53*10^6 , 169^(2.88)=2.59*10^6 , 6.53*10^6 not 2.59*10^6 , false ,

289^x - 221^x = 169^x 17²^x - 13^x*17^x = 13²^x put, a = 17^x , b = 13^x a² - ab - b² = 0 a = [-(-b) ± √(-b)² - 4*1*(- b²) ] / 2*1 a = b ± √5*b) / 2 a = b/2 * (1 ± √5) a = b/2 * (1 + √5) 17^x = 13^x/2 * (1 + √5) 17^x / 13^x = (1 + √5)/2 (17 / 13)^x = (1 + √5)/2 log(17 / 13)^x = log(1 + √5)/2 x * log(17 / 13) = log(1 + √5)/2 x = log(1 + √5)/2 / log(17 / 13) x = [log(1 + √5) - log2] / (log17 - log13)

Good Sir

(17^2)^x-(13*17)^x=(13^2)^x , /:(13^2)^x , (17/13)^(2x)-(17/13)^(x)-1=0 , let u=(17/13)^x , u^2-u-1=0 , u=(1+V5)/2 , /(1-V5)/2 < 0 , not a solu / , x=log((1+V5)/2)/log(17/13) , x=~ 1.7938 , test , 289^x-221^x=~ 9917.04 , 169^x=~ 9917.04 , OK ,