log maths question

1) (log a + 2 log b)/[log (ab) - log √a]

= (log a + 2 log b)/[log a + log b - (1/2) log a]

= (log a + 2 log b)/[(1/2) log a + log b]

= 2

2) [(1/3) log (x2y3) - log (x3y)]/[log (x1/3)]

= [(2/3) log x + log y - 3 log x - log y]/[(1/3) log x]

= [(-4/3) log x]/[(1/3) log x]

= -4

3) [log (√x y3/2) - log (xy)]/[(-3/2) log (xy) + log (x2y)]

= [(1/2) log x + (3/2) log y - log x - log y]/[(-3/2) log x - (3/2) log y + 2 log x + log y]

= [-(1/2) log x + (1/2) log y]/[(1/2) log x - (1/2) log y]

= -1

2011-08-11 17:28:06 補充：
[-(1/2) log x + (1/2) log y]/[(1/2) log x - (1/2) log y]
= - [(1/2) log x - (1/2) log y]/[(1/2) log x - (1/2) log y]
= -1

(log a + 2 log b)/[(1/2) log a + log b]
= 2[(1/2) log a + log b]/[(1/2) log a + log b]
= 2

2011-08-11 20:16:08 補充：
(loga+2logb) loga都冇2點抽??

照抽 2 出黎, 跟住入面乘返 1/2 作為抵銷.