Question

16^(x+1/4)-41*4^(x-1)+9=0

Answer 1

$16^{x+frac{1}{4}}-41*4^{x-1}+9=0\ 8*4^{2x}-41*4^x+36=0\ 4^x=t\ 8t^2-41t+36=0\ D=23^2\ t=frac{41+23}{16}=4\ 4^x=4\ x=1$