證明 log7^11+log7^29/2<log7^20?

Answer 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)