有啲數學既問題唔識做...唔該入黎教教我 - 20分!

1)3^3(y+1) = 7^y x 5^(y+3)
27^(y+1) = 7^y*5^y*5^3
27*27^y = 7^y*5^y*125
27^y/(7^y*5^y)=125/27
[27/(7*5)]^y=125/27
log[27/(7*5)]^y=log125/27
ylog(27/35) = log125/27 (logM^y=ylogM)
y = ( log125/27)/log(27/35)
y =-5.9052(4位小數)

2) 10^(3y-5) = 2^(2y-7)
兩面take log:
(3y-5)log10=(2y-7)log2 ( log10=1)
3y-5=y*2log2 - 7log2
y(3-2log2)=5 - 7log2
y=(5-7log2)/(3-log4)
y=1.2064(4位小數)

2009-01-03 20:27:44 補充：
3樓的id0081406,你做乜要抄我答案?!你的行為已經構成侵犯知識產權的罪行,並且違反了知識+的規章,嚴重浪費了網絡空間。我會保留一切法律追究權利。望你好自為之!

1) 3^3(y+1) = 7^y x 5^(y+3)
27^(y+1)=7^y*5^y*5^3
27*27^y = 7^y*5^y*125
27^y/(7^y*5^y)=125/27
[27/(7*5)]^y=125/27
log[27/(7*5)]^y=log125/27
ylog(27/35) = log125/27 (logM^y=ylogM)
y = ( log125/27)/log(27/35)
y =-5.9052(4位小數)

2) 10^(3y-5) = 2^(2y-7)
兩面take log:
(3y-5)log10=(2y-7)log2 ( log10=1)
3y-5=y*2log2 - 7log2
y(3-2log2)=5 - 7log2
y=(5-7log2)/(3-log4)
y=1.2064(4位小數)

1.
3^(3y+1) = (7^y)[5^(y+3)]
(3)[3^(3y)] = (7^y)(5^3][5^y]
(3)(27^y) = 125(35)^y
(27/35)^y = 125/3
y log(27/35) = log(125/3)
y = log(125/3)/log(27/35) = 1.619789/0.77143 =2.1
2.
10^(3y-5) = 2^(2y-7)
10^(3y)/10^5 = 2^(2y)/2^7
1000^y/10^5 = 4^y/128
(1000/4)^y = 10^5/128
250^y = 10^5/128
y log250 = log(10^5/128)
y = log(10^5/128)/log 250 = 2.898/2.398 = 1.2085.



2009-01-03 18:27:34 補充：
For Q1. 3^3(y +1) should be (27)(3^y), not (3)(3^y) Sorry for the mistake.