how do I solve this?
logs?
5^(x) = 4^(2x+1) ?????
2009-03-06 9:58 am
回答 (4)
2009-03-06 11:01 am
✔ 最佳答案
5^x = 4^(2x + 1)log(5^x) = log[4^(2x + 1)]
(x)log(5) = (2x + 1)log(4)
(x)log(5) = (2x)log(4) + log(4)
(x)log(5) - (2x)log(4) = log(4)
x[log(5) - 2log(4)] = log(4)
x = log(4)/[log(5) - 2log(4)]
2009-03-06 10:10 am
x ln5 = (2x+1) ln4
x ln5 = 2xln4 + ln4
x ln5 = xln16 + ln4
x = ln4 / (ln5 - ln16)
= -1.192
x ln5 = 2xln4 + ln4
x ln5 = xln16 + ln4
x = ln4 / (ln5 - ln16)
= -1.192
2009-03-06 10:07 am
Don't Know maybe 12
2009-03-06 10:05 am
x log 5 = (2x + 1) log 4
x log 5 = (log 4) (2) x + log 4
( log 5 - 2 log 4 ) x = log 4
x = log 4 / (log 5 - 2 log 4)
x log 5 = (log 4) (2) x + log 4
( log 5 - 2 log 4 ) x = log 4
x = log 4 / (log 5 - 2 log 4)
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