How to do the following qu?
1.(5^(x+2))(7^x)=3
2.
log(x-5y)=2______(1)
logx-logy=1______(2)
Pls show steps clearly. Thank you!
F4 math log
2011-01-31 3:54 am
回答 (3)
2011-01-31 4:48 am
2011-01-31 11:01 pm
1. (5^(x+2))(7^x)=3
(5^x × 5^2)(7^x)=3
25(5^x)(7^x)=3
(5^x)(7^x)=0.12
㏒[(5^x)(7^x)]=㏒0.12
x㏒5 + x㏒7=㏒0.12
x(㏒5+ ㏒7)=㏒0.12
x=㏒0.12/(㏒5+ ㏒7)
x=-0.59636
2.
log(x-5y)=2______(1)
logx-logy=1______(2)
From(2),logx-logy=1
log(x÷y)=1
(x÷y)=10
x=10y______(3)
Subst. (3) into (1),
log(10y-5y)=2
log(5y)=2
5y=10^2
y=20
Subst. y=20 into (3),
x=10(20)
x=200
2011-01-31 15:02:30 補充：
100% correct~
(5^x × 5^2)(7^x)=3
25(5^x)(7^x)=3
(5^x)(7^x)=0.12
㏒[(5^x)(7^x)]=㏒0.12
x㏒5 + x㏒7=㏒0.12
x(㏒5+ ㏒7)=㏒0.12
x=㏒0.12/(㏒5+ ㏒7)
x=-0.59636
2.
log(x-5y)=2______(1)
logx-logy=1______(2)
From(2),logx-logy=1
log(x÷y)=1
(x÷y)=10
x=10y______(3)
Subst. (3) into (1),
log(10y-5y)=2
log(5y)=2
5y=10^2
y=20
Subst. y=20 into (3),
x=10(20)
x=200
2011-01-31 15:02:30 補充：
100% correct~
2011-01-31 4:39 am
2. log (x - 5y)= 2 _______(1)
logx - logy= 1_______ (2)
logx - log5y= 2
log(x/5y)= 2
x/5y= 2 _________(1a)
logx - logy= 1
log(x/y)= 1
x/y=1
x=y____________(2a)
Substitute x=y into 1a
y/5y=2
y=10y
y=0
Substitute y= 0 into 1a again,
x/5(0)=2
x=2
logx - logy= 1_______ (2)
logx - log5y= 2
log(x/5y)= 2
x/5y= 2 _________(1a)
logx - logy= 1
log(x/y)= 1
x/y=1
x=y____________(2a)
Substitute x=y into 1a
y/5y=2
y=10y
y=0
Substitute y= 0 into 1a again,
x/5(0)=2
x=2
參考: My mathematics ability
收錄日期: 2021-04-24 10:07:01
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110130000051KK01011