Question

sabendo 2^x/2^-x=5, o valor de 8^x+ 8^-x e igual ?

Answer 1

$\displaystyle \frac{2^{x}}{2^{-x}}=5 \\ \\ \\ \displaystyle \frac{2^{x}}{\displaystyle \frac{1}{2^{x}}}=5 \\ \\ \\ 2^{x} \cdot 2^{x} =5 \\ \\ \\ 4^{x}=5 \\ \\ \\ \ln4^{x}=\ln5 \\ \\ \\ x \cdot \ln4=\ln5 \\ \\ \\ x=\frac{\ln5}{\ln4} \\ \\ \\ x \approx 1,16$

Se x ≅ 1,16, então:

$\displaystyle 8^{x}+8^{-x} \\ \\ \\ 8^{x} + \frac{1}{8^{x}} \\ \\ \\ 8^{1,16}+ \frac{1}{8^{1,16}} \\ \\ \\ \boxed{\boxed{ \approx 11,25}}$