點解 log x^k= k log x

Prove by Mathematic induction .

when n=1 , log x^(1) = (1) log x , it is true for n=1 ,

Assume it is true for n= k , where k belongs to nature number ,

i.e. log x^k = k log x

Consider n=k+1 , log x^(k+1) = log x^k x = log x^k + log x = (k+1)logx

it is true for n=k+1 ,

By Mathematical induction , it is true for all positive integer n.

For n=0 , log x^0 = log 1 = 0 = (0) log x

it is ture all positive integers.

For negetivei integer , let n=-m ,

log x^n = log x^-m = log 1/x^m = log 1- log x^m =- log x^m = -m log x

it is true for all integers.



2009-07-31 23:20:46 補充：
For rational number ,

let n=a/b , where a and b are integer and no common factor .

by the above result , we know that log x^a=a log x and log x^b = blogx ,

2009-07-31 23:23:40 補充：
Consider blogx^n = log x^(nb) = log x^a = alogx

i.e. log x^n = (a/b)logx = nlogx

so it is true for all rational numbers .

log(-2)^4=4log(-2)

定理 : log a*b = log a+ log b
log a*b*c = log a+ log b+ log c
log a*a =" log a^2 "= log a+ log a=" 2 log a "

log x^k
= log x*x*x*x*...x (共乘k次) = log x+ log x+ ... log x (加k次)
= k log x

你知道log點解嗎？假設：
logX Y = Z；
XZ = Y；
X的Z次方等於Y，這是log的簡單解釋。
假如：
XZ = Y；
雙邊都以power of k來剩，即：
XZ(k)= Yk；
亦即是：
k (logX Y)； 或
k(Z)