Question

16^(x)+16^(-x)=47

Answer 1

\[16^{x}+16^{-x}=47\]

Answer 2

\(\Large 16^{-x}=\frac{1}{16^x}\)
thus, \(\Large 16^{x}+16^{-x}=47\Leftrightarrow 16^x+\frac{1}{16^x}-47=0\)

Let \(\Large t=16^x\), Therefore:
\(\Large t+\frac{1}{t}-47=0\Rightarrow t^2-47t+1=0\)

\(\Large \Rightarrow t=\frac{47\pm21\sqrt{5}}{2}\)

Hence, \(\Large x=\log _{ 16 }{( t) } \)