How do I solve using Log?
(1/9)^m=81^m+4
How do I solve this equation?
2009-04-06 6:33 am
回答 (6)
2009-04-06 6:43 am
✔ 最佳答案
Take log of both sidesSo m log(1/9) = (m+4)log81
-m log9 = 2(m+4)log9......Using laws of logarithms.
So -m = 2m + 8 or m = - 8/3
2009-04-06 2:02 pm
(1/9)^m=81^m+4
Take log on both sides, we have
log(1/9)^m=log{81^(m+4)}
or m log1/9)=(m+4)log 81
m (log1- log9)=(m+4)log 9^2
or m (0 - log9)=2(m+4)log 9
or -m log9=2mlog 9 +8log9
or -m log9-2m log9 =8log9
-3m log9 =8log9
m = - 8/3
Take log on both sides, we have
log(1/9)^m=log{81^(m+4)}
or m log1/9)=(m+4)log 81
m (log1- log9)=(m+4)log 9^2
or m (0 - log9)=2(m+4)log 9
or -m log9=2mlog 9 +8log9
or -m log9-2m log9 =8log9
-3m log9 =8log9
m = - 8/3
2009-04-06 1:47 pm
9^-m=9^(2m+8)
m=-8/3
m=-8/3
2009-04-06 1:44 pm
(1/9)^m = 81^(m + 4)
(9^-1)^m = (9^2)^(m + 4)
9^(-m) = 9^[2(m + 4)]
log_9[9^(-m)] = log_9{9^[2(m + 4)]}
-m = 2(m + 4)
-m = 2m + 8
-m - 2m = 8
m = 8/-3
m = -8/3
(9^-1)^m = (9^2)^(m + 4)
9^(-m) = 9^[2(m + 4)]
log_9[9^(-m)] = log_9{9^[2(m + 4)]}
-m = 2(m + 4)
-m = 2m + 8
-m - 2m = 8
m = 8/-3
m = -8/3
2009-04-06 1:43 pm
ln(1/9)^m = ln(81)^m +ln4
using the rules of logs ln(a)^b = blna
m.ln(1/9)=m.ln(81) + ln4
m( ln(1/9) - ln(81) = ln 4
m = ln4 / ( ln(1/9) - ln(81) )
using rules of logs lna - lnb = ln(a/b)
m = ln4 / ln( (1/9)/81 )
using the rules of logs ln(a)^b = blna
m.ln(1/9)=m.ln(81) + ln4
m( ln(1/9) - ln(81) = ln 4
m = ln4 / ( ln(1/9) - ln(81) )
using rules of logs lna - lnb = ln(a/b)
m = ln4 / ln( (1/9)/81 )
2009-04-06 1:42 pm
first, subtract the (1/9)^m from the 81^m to get:
(728/9)^m=-4
Then, using log, we get:
log(base 728/9) -4 = m
(728/9)^m=-4
Then, using log, we get:
log(base 728/9) -4 = m
收錄日期: 2021-05-01 12:14:36
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