I was just wondering if i can do this.
Q. Solve $\log_{9}24=x $
$\implies9^x =24$
$\implies3^{2x}=2^3 3$
$\implies\log_3(3^{2x})= \log_3(2^3 3)$
$\implies2x=2 (3)^{1/3}$
$\implies x=3^{1/3} $
Is this actually correct or did i break some kind of log rule here? My answer appears to be out by 0.0038 compared to the book is that cause they used a calculator or is it just a fluke that my answer is so close?
Your method is incorrect as pointed out by Mufasa (sorry I didn't spot that earlier) but it's not the way I'd have done it anyway (and probably not how the book answer was obtained). This way is better:
$9^x = 24$
$\implies x\ln 9 = \ln 24$
$\implies x = \frac{\ln 24}{\ln 9} = 1.44639...$