f.5對數函數1題

First of all, change log 729 to the base x to log 729/log x both to the base 3.
log 729 = log 3^6 = 6 log 3 to the base 3 = 6.
Let log x to the base 3 = y
2y - 6/y = 1
2y^2 - y - 6 = 0
(2y + 3)(y - 2) = 0
y = - 3/2 or 2
that is log x to the base 3 = - 3/2 or 2
so x = 3^(-3/2) or 3^2
= 1/(sqrt 27) or 9.

2012-01-09 21:18:40 補充：
log x^2 = 2 log x = 2y. log 729/log x = 6/y, so the equation becomes 2y - 6/y = 1.

圖片參考：http://imgcld.yimg.com/8/n/HA00285646/o/701201090072513873420010.jpg


右鍵另存會放大D
2個方法都得, 第2個就正常人做法, 見到有BASE轉哂做10, 禁到機
第1個就係勁人做法, 一睇就知轉BASE 3, 因為729係3^k次

轉base(底)公式 係 log{a} ( b) = log(b)/ log(a)
如果你唔知有呢樣野就做唔到條數

最後個log x, 為左方便計, 會用y=log(x)表示
計到y姐係計到x

2012-01-09 23:27:53 補充：
答案寫3位有效數字就得, 唔洗咩1/sqrt(27)

設 u = logx
log3x^2 - logx 729 = 1
(2u)/log 3 - (log729)/u = 1
2u^2 - 3 (log 3 )^2 = (log 3)u
(2u - 3log 3)(u + log 3)=0
so u=(3/2)log3 or u = -log3
so x= 開方27 or x=1/3
remarks: log x^2 = 2log x
log3 x^2= (logx^2)/log3