(log 24)^2 + (log3)^2-(log24)(log9)=?
.....2................2............2.......2......=>(以2為底)
註明:
^2 =>表示2次方
求詳解及過程
謝謝~
奉上20點!很急~要考試了! log對數的計算問題求解?
2014-01-03 3:53 am
回答 (3)
2014-01-03 4:06 am
✔ 最佳答案
(log2_24)^2+(log2_3)^2－(log2_24)(log2_9)=?Sol
A=(log2_24)^2+(log2_3)^2－(log2_24)(log2_9)
A(log2)^2
=(log24)^2+(log3)^2－log24*log9
=(log24)^2+(log3)^2－2log24*log3
=(log24－log3)^2
=(log8)^2
=(3log2)^2
=9(log2)^2
2014-01-04 8:38 pm
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俆僔
俆僔
2014-01-03 4:11 am
以下log皆以2為底
(log 24)^2 + (log3)^2 -(log24)(log9)
= (log 24)^2 -(log24)(log3^2) + (log3)^2
= (log 24)^2 -2(log24)(log3) + (log3)^2
= ( log24 - log3 )^2
= ( log2^3*3 - log3 )^2
= ( 3 + log3 - log3 )^2
= 3^2
= 9
Ans: 9
(log 24)^2 + (log3)^2 -(log24)(log9)
= (log 24)^2 -(log24)(log3^2) + (log3)^2
= (log 24)^2 -2(log24)(log3) + (log3)^2
= ( log24 - log3 )^2
= ( log2^3*3 - log3 )^2
= ( 3 + log3 - log3 )^2
= 3^2
= 9
Ans: 9
收錄日期: 2021-04-30 18:21:10
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140102000015KK03901