9 = 40(2)^2x+1
I think he wants,
$\displaystyle 9 = 40(2)^{2x+1 } $
In which case we use basic logs.
$\displaystyle ln ( \frac{ 9}{40} ) = ln (2^{2x+1}) = (2x+1) ln2 $
$\displaystyle \frac { ln ( \frac{ 9}{40} ) }{ ln2} = 2x + 1 $
$\displaystyle \frac { ln ( \frac{ 9}{40} ) }{2 ln2} - \frac{1}{2} = x $