Solve the equation 7^(x – 2) = 5^x. Round the answer to four decimal places?
回答 (4)
2010-06-18 4:23 am
✔ 最佳答案
When your variable is in the exponent position consider taking the log of both sides.log 7^(x-2) = log 5^x
(x-2)* log 7 = x * log 5
x*log 7 - 2*log 7 = x * log 5
x log 7 - x log 5 = 2 log 7
x (log7 - log5 ) = 2 log 7
x = (2 log 7) / (log 7 - log 5)
x = 11.5665
2010-06-18 7:44 am
( x - 2 ) log 7 = x log 5
x log 7 - x log 5 = 2 log 7
x [ log 7 - log 5 ] = 2 log 7
x = 2 log 7 / [ log (7/5) ]
x = 11.5665
x log 7 - x log 5 = 2 log 7
x [ log 7 - log 5 ] = 2 log 7
x = 2 log 7 / [ log (7/5) ]
x = 11.5665
2010-06-18 5:05 am
7^(x - 2) = 5^x
log[7^(x - 2)] = log(5^x)
(x - 2)log(7) = (x)log(5)
(x - 2)/x = log(5) / log(7)
1 - 2/x = log(5) / log(7)
2/x = 1 - log(5) / log(7)
x = 2/[1 - log(5) / log(7)]
x = 2/[log(7) / log(7) - log(5) / log(7)]
x = 2/[log(7) - log(5) / log(7)]
x = 2[log(7)] / [log(7) - log(5)]
x = 2log(7) / [log(7) - log(5)]
x = 11.5665421
log[7^(x - 2)] = log(5^x)
(x - 2)log(7) = (x)log(5)
(x - 2)/x = log(5) / log(7)
1 - 2/x = log(5) / log(7)
2/x = 1 - log(5) / log(7)
x = 2/[1 - log(5) / log(7)]
x = 2/[log(7) / log(7) - log(5) / log(7)]
x = 2/[log(7) - log(5) / log(7)]
x = 2[log(7)] / [log(7) - log(5)]
x = 2log(7) / [log(7) - log(5)]
x = 11.5665421
參考: 2 log 7 / (log 7 - log 5) - Google Search
http://www.google.com/search?hl=en&q=2+log+7+/+(log+7+-+log+5)&aq=f&aqi=h1&aql=&oq=&gs_rfai=
2010-06-18 4:17 am
7^(x - 2) = 5^x
(7^x)(7^(-2)) = 5^x
(7^x)/(7^2) = 5^x
(7^x)/49 = 5^x
1/49 = (5^x)/(7^x)
49 = (7^x)/(5^x)
49 = (7/5)^x
log[7/5](49) = x
x ≈ 11.5665
(7^x)(7^(-2)) = 5^x
(7^x)/(7^2) = 5^x
(7^x)/49 = 5^x
1/49 = (5^x)/(7^x)
49 = (7^x)/(5^x)
49 = (7/5)^x
log[7/5](49) = x
x ≈ 11.5665
收錄日期: 2021-05-01 13:14:48
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100617201253AAEyYd2