Решение логарифмического выражения.

Решите задачу:

$\frac{log2(24)- \frac{1}{2}log2(72) }{log3(18)-\frac{1}{3}log3(72)}=\frac{\frac{2log2(24)-log2(72)}{2}}{\frac{3log3(18)-log3(72)}{3}}=\frac{(2log2(24)-log2(72))*3}{2(3log3(18)-log3(72))}= \\ =\frac{6log2(24)-3log2(72)}{2(log3(18^3)-log3(72))}=\frac{log2(24^6*72^{-3}) }{2log3(\frac{18^3}{4*18})} =\frac{log2((2^3)^6*3^6*3 ^{-3}*3 ^{-3} (2^3)^{-3} }{2log3((\frac{18}{2})^2)}= \\ =\frac{log2(2^9)}{4log3(\frac{18}{2})}=\frac{log2(2^9)}{4log3(3^2)}=\frac{9}{4*2}=\frac{9}{8}=1,125$