20. URCA (2013.2) - Se x=log35/2 . log53/3 . log342.log253 então, 2^(1/x) vale: a) 2 b) 2√2 c) 2 3√4 d) 9√25 e) 8 3√2

Para resolver essa questão, vamos utilizar as propriedades dos logaritmos. Começando por x=log35/2 . log53/3 . log342.log253, podemos reescrever cada logaritmo como uma fração: x = log2(35)/log2(2) . log3(53)/log2(3) . log2(342)/log2(10) . log2(253)/log2(10) Podemos simplificar os dois últimos logaritmos: x = log2(35)/log2(2) . log3(53)/log2(3) . log2(342.253)/log2(10^2) x = log2(35)/log2(2) . log3(53)/log2(3) . log2(86406)/log2(100) x = log2(35)/log2(2) . log3(53)/log2(3) . log2(86406)/2 x = (log2(35) . log3(53) . log2(86406))/2(log2(2) . log2(3) . log2(100)) x = (log2(35) . log3(53) . log2(86406))/2(log2(2.3.10^2)) x = (log2(35) . log3(53) . log2(86406))/2(log2(600)) x = (log2(35) . log3(53) . log2(86406))/2(log2(2.2.3.5.5)) x = (log2(35) . log3(53) . log2(86406))/(log2(2^3.3.5)) x = (log2(35) . log3(53) . log2(86406))/(3log2(2) + log2(3) + log2(5)) x = (log2(35) . log3(53) . log2(86406))/(3.1 + log2(3) + log2(5)) x = (log2(35) . log3(53) . log2(86406))/(4 + log2(3) + log2(5)) Agora, podemos substituir o valor de x na expressão 2^(1/x): 2^(1/x) = 2^(2(4+log2(3)+log2(5))/(log2(35) . log3(53) . log2(86406))) 2^(1/x) = (2^4 . 2^(log2(3)) . 2^(log2(5)))^(2/(log2(35) . log3(53) . log2(86406))) 2^(1/x) = (16 . 3 . 5)^(2/(log2(35) . log3(53) . log2(86406))) 2^(1/x) = 240^(2/(log2(35) . log3(53) . log2(86406))) 2^(1/x) = 240^(2/((log10(35)/log10(2)) . (log10(53)/log10(3)) . (log10(86406)/log10(2)))) 2^(1/x) = 240^(2/((1.543/1) . (1.727/0.477) . (4.936/1))) 2^(1/x) = 240^(2/(3.238)) 2^(1/x) = 240^(0.617) 2^(1/x) = 8.3 Portanto, a alternativa correta é a letra E) 8 3√2.