更新1:
okay.... after 5 minutes of reviewing that made sense...
log(base 7)343 = 5x - 7?
回答 (6)
2009-03-18 9:43 am
✔ 最佳答案
log(base 7)343 = 5x - 7log(base 7)7³ = 5x - 7
3log(base 7)7 = 5x - 7
3 = 5x - 7
3 + 7 = 5x
10 = 5x
x = 2
參考: Common Exponential and Log Identities: http://www.karlscalculus.org/explogid.html
2009-03-18 5:08 pm
log_7(343) = 5x - 7
343 = 7^(5x - 7)
7^3 = 7^(5x - 7)
3 = 5x - 7
5x = 3 + 7
x = 10/5
x = 2
343 = 7^(5x - 7)
7^3 = 7^(5x - 7)
3 = 5x - 7
5x = 3 + 7
x = 10/5
x = 2
2009-03-18 4:56 pm
ok...it goes like this..
log (base 7) 343 = 5x - 7 --. since 7 x 7 x 7 = 343...
3 = 5x - 7
10 = 5x
x = 2
hope it will be of some help! :)..all the best!
log (base 7) 343 = 5x - 7 --. since 7 x 7 x 7 = 343...
3 = 5x - 7
10 = 5x
x = 2
hope it will be of some help! :)..all the best!
2009-03-18 4:38 pm
x=2
2009-03-18 4:36 pm
log(base 7)7^3 = 5x - 7
3log(base 7)7 = 5x - 7
3 = 5x - 7
10 = 5x
x = 2
remember log(base a)a = 1
3log(base 7)7 = 5x - 7
3 = 5x - 7
10 = 5x
x = 2
remember log(base a)a = 1
2009-03-18 4:35 pm
Since 343=7^3
log(base 7)343=3
So 5x=10 and x=2
log(base 7)343=3
So 5x=10 and x=2
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090318013026AAtlPQF