logarithmic functions

ab = 27
log (ab) = log27
log a + log b = 3log3 ... (1)
log² a + log² b + 2 (log a) log b = 9 log² 3 ... (2) loga b + logb a = 2.5
log b / log a + log a / log b = 2.5
log² a + log² b = 2.5 (log a) log b ... (3) Substitute (3) into (2) :4.5 (log a) log b = 9 log² 3
(log a) log b = 2 log² 3 ... (4) (2) - (4)*4 :
log² a + log² b - 2 (log a) log b = 9 log² 3 - 8 log² 3
(log a - log b)² = log² 3
log a - log b = ± log 3 ... (5) (1) + (5) :
2 log a = 3 log 3 ± log 3
log a = 2 log 3 = log 9 or log a = log 3
so
log b = log 3 or log b = 2 log 3 = log 9Therefore (a , b) = (3 , 9) or (a , b) = (9 , 3)

2012-07-31 23:14:55 補充：
Method 2 :

Let A = loga(b) , then 1/A = logb(a).

A + 1/A = 2.5
A = 2

So log a : log b = 2 : 1 or 1 : 2

then a = b² or b = a² ,
(a = 3 , b = 9) or (a = 9 , b = 3) since ab = 27.

I think this is a better way.