更新1:
why can (100^log2)+(100^log3) =2^log100 + 3^log100
(100^log2)+(100^log3) help
回答 (5)
2008-06-16 3:57 am
✔ 最佳答案
題數你睇上面兩位已經知點計我就答你點解可以咁做
其實第一位答既跳左唔少步驟
留意10^loga=a
因為如果loga=b,咁10^b就等於a
我當log4=a,咁100^log2=10^2log2=10^1og4=10^a姐係 4啦
＊loga=b,10^b=a
我相信e度係比較難明
P.S:張遼唔係話唔答問題咩？
2008-06-16 6:22 am
001: 請問你d步驟點嚟ka? 既然識得話人跳步, 咁點解自己又會嘅?
http://hk.knowledge.yahoo.com/question/question?qid=7008061501919
http://hk.knowledge.yahoo.com/question/question?qid=7008061501919
2008-06-15 11:37 pm
100^(log2) + 100^(log3)
= (10^2)^(log2) + (10^2)^(log3)
= 10^(2log2) + 10^(2log3)
= 10^[log(2^2)] + 10^[log(3^2)]
= 10^(log 4) + 10^(log 9)
= 4 + 9
= 13
= (10^2)^(log2) + (10^2)^(log3)
= 10^(2log2) + 10^(2log3)
= 10^[log(2^2)] + 10^[log(3^2)]
= 10^(log 4) + 10^(log 9)
= 4 + 9
= 13
2008-06-15 11:26 pm
(100^log2)+(100^log3)
=2^log100 + 3^log100
=4+9
=13
2008-06-16 20:15:53 補充：
我以為有這一條公式,恩....
a^logb=b^loga
a^logb=10^[log(a^logb)]
=10^[logb loga]
=(10^logb)^loga .... r^(ab)=(r^a)^b [ 2^(2x3) = (2^2)^3
=b^loga .....10^logb=b Simularly, In (e^b) = b
=2^log100 + 3^log100
=4+9
=13
2008-06-16 20:15:53 補充：
我以為有這一條公式,恩....
a^logb=b^loga
a^logb=10^[log(a^logb)]
=10^[logb loga]
=(10^logb)^loga .... r^(ab)=(r^a)^b [ 2^(2x3) = (2^2)^3
=b^loga .....10^logb=b Simularly, In (e^b) = b
參考: myself
收錄日期: 2021-04-26 20:25:15
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080615000051KK01362