[Equação exponencial]

por **JU201015** » Dom Nov 11, 2012 19:22

Se ${2}^{x}+{2}^{-x}=3$ , o valor de ${8}^{x}+{8}^{-x}$ é?

por **DanielFerreira** » Dom Nov 11, 2012 19:58

por **JU201015** » Dom Nov 11, 2012 22:33

danjr5 escreveu: $\\ 2^x + 2^{- x} = 3 \\\\ \left ( 2^x + \frac{1}{2^x} \right ) = 3 \\\\\\ \left ( 2^x + \frac{1}{2^x} \right )^3 = 3^3 \\\\\\ 2^{3x} + 3 \cdot 2^{2x} \cdot \frac{1}{2^x} + 3 \cdot 2^{x} \cdot \frac{1}{2^{2x}} + \frac{1}{2^{3x}} = 27 \\\\\\ 2^{3x} + 3 \cdot 2^x + 3 \cdot \frac{1}{2^x} + \frac{1}{2^{3x}} = 27 \\\\\\ 2^{3x} + 3 \left ( 2^x + \frac{1}{2^x} \right ) + \frac{1}{2^{3x}} = 27 \\\\\\ (2^3)^x + 3 \cdot 3 + \frac{1}{(2^3)^{x}} = 27 \\\\\\ 8^x + 9 + \frac{1}{8^{x}} = 27 \\\\\\ 8^x + 8^{- x} = 27 - 9 \\\\\\ \boxed{8^x + 8^{- x} = 18}$

Muito obrigada! Entendi bem. Mas, esse negócio de elevar ao cubo eu nunca pensaria =s kk