A [m(3^y)]/(nlog9)
B (ylogm)/(logn)
C 1/2 [(logm-logn)/log3 +y ]
D (m+3^y)/n
E m(3^y) -n
更新1:
要步驟
Maths!!!
2008-01-01 7:38 am
已知9^x /3^y =m/n ,試以m、n和y表示x。
A [m(3^y)]/(nlog9)
B (ylogm)/(logn)
C 1/2 [(logm-logn)/log3 +y ]
D (m+3^y)/n
E m(3^y) -n
A [m(3^y)]/(nlog9)
B (ylogm)/(logn)
C 1/2 [(logm-logn)/log3 +y ]
D (m+3^y)/n
E m(3^y) -n
回答 (3)
2008-01-01 7:43 am
✔ 最佳答案
9^x /3^y =m/n3^(2x-y)=m/n
(2x-y)log3=log(m/n)
(2x-y)log3=logm-logn
2x-y=(logm-logn)/log3
∴x=1/2 [(logm-logn)/log3 +y ]
C is the answer
2008-01-01 7:54 am
answer is C
9^x/3^y=m/n
3^(2x-y)=m/n
log3^(2x-y)=log(m/n)
(2x-y)log3=logm-logn
y=1/2 [(logm-logn)/log3 +y ]
9^x/3^y=m/n
3^(2x-y)=m/n
log3^(2x-y)=log(m/n)
(2x-y)log3=logm-logn
y=1/2 [(logm-logn)/log3 +y ]
2008-01-01 7:49 am
The correct answer is C
9^x/3^y=m/n
3^(2x-y)=m/n
log3^(2x-y)=log(m/n)
(2x-y)log3=logm-logn
y=1/2 [(logm-logn)/log3 +y ]
9^x/3^y=m/n
3^(2x-y)=m/n
log3^(2x-y)=log(m/n)
(2x-y)log3=logm-logn
y=1/2 [(logm-logn)/log3 +y ]
收錄日期: 2021-04-23 20:32:45
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071231000051KK04738