If $1+2x+3x^2 + \dotsb=9$, find $x$.
Let $S = 1 + 2x + 3x^2 + \dotsb.$  Then
\[xS = x + 2x^2 + 3x^3 + \dotsb.\]Subtracting these equations, we get
\[(1 - x) S = 1 + x + x^2 + \dotsb = \frac{1}{1 - x},\]so $S = \frac{1}{(1 - x)^2}.$  Thus, we want to solve
\[\frac{1}{(1 - x)^2}  = 9.\]then $(1 - x)^2 = \frac{1}{9},$ so $1 - x = \pm \frac{1}{3}.$  Since $x$ must be less than 1, $1 - x = \frac{1}{3},$ so $x = \boxed{\frac{2}{3}}.$