logarithm

darksider

为什么 a^loga N = N?

3^log3 7=?   3是小型的

8^log2 5=?   2是小型的

有人可以解释吗？

xanadunext

consider a^loga N = X

loga (a^loga N) = loga X

(loga N)(loga A) = loga X

loga N ( 1 ) = loga X

N = X

therefore a^loga N = N

darksider

谢谢。。。。。。。。

darksider

2^a=5^b=10^c

Explain a in terms of b and c

thanks

TCLGT

原帖由 <i>darksider</i> 于 2-8-2007 08:10 PM 发表 <a href="http://chinese.cari.com.my/myforum/redirect.php?goto=findpost&pid=31051447&ptid=937619" target="_blank"><img src="http://chinese.cari.com.my/myforum/images/common/back.gif" border="0" onload="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onmouseover="if(this.width>screen.width*0.7) {this.resized=true; this.width=screen.width*0.7; this.style.cursor='hand'; this.alt='Click here to open new window\nCTRL+Mouse wheel to zoom in/out';}" onclick="if(!this.resized) {return true;} else {window.open(this.src);}" onmousewheel="return imgzoom(this);" alt="" /></a><br />
2^a=5^b=10^c<br />
<br />
Explain a in terms of b and c<br />
<br />
thanks

<br />

a log 2= b log 5= c log 10

let ,a log 2= d , log2 = d/a
     b log 5=d, log 5=d/b
     c log 10=d, log 10=d/c

since log 2+log5= log10
d/a+d/b=d/c

1/a+1/b=1/c
1/a=1/c-1/b
a=1/(1/c-1/b)
a=bc/(b-c)

mathlim

Proof without logarithm：

2^a = 5^b = 10^c

2^a = 10^c
2^ab = 10^bc ——(1)

5^b = 10^c
5^ab = 10^ca ——(2)

(1)×(2)
10^ab = 10^(bc+ca)
ab = bc + ca
ab - ca = bc
a(b-c) = bc
a = bc/(b-c)

mathlim

a^logaN = N

Let M = logaN
a^M = N
a^logaN = N