Need some help on these, just to make sure I understand what I'm doing...

1) If $\displaystyle \log_ba = x$ then $\displaystyle \log_ab =$

$\displaystyle \frac{1}{\log_ab}$

2) If $\displaystyle \log_ba = x$ then $\displaystyle \log_\frac{1}{b}a =$

$\displaystyle -\log_ba$

3) $\displaystyle \log(2 - x) + \log(3 - x) = \log12$ solve for x

$\displaystyle x = -1$

4) $\displaystyle y = \log_bx + \log_b(x - 6) $ solve for x

$\displaystyle b^y = x^2 - 6x$

then...

would the quadratic formula be used?

$\displaystyle x^2 - 6x - b^y = 0$

??