證明 log7^11+log7^29/2<log7^20?
回答 (1)
What is "log7^11+log7^29/2" ?
若 log 的底大於 1. 則
(log7^11)+(log(7^29))/2
= log(7^11) + log((7^29)^(1/2))
= (log7^11)+(log7^(29/2))
= log((7^11)(7^14.5))
= log(7^25.5) > log(7^20)
(log7^11)+(log((7^29)/2))
> log(7^11) + log(7^28)
= log(7^39) > log(7^20)
可能造成 <log7^20 的, 恐怕只有
log(7^11+log7^29/2)?
(log(7^29))/2 = log7^(29/2) < log((7^29)/2)
log(7^11+log7^29/2)
≦ log(7^11+ log((7^29)/2))
< log(7^11+ log(7^29))
< log(7^11+ 29) (假設 log 的底是 10)
< log(7^11 + 7^2)
< log((7^11)×2)
< log(7^12)
< log(7^20)
收錄日期: 2021-05-04 02:36:19
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