證明 log7^11+log7^29/2<log7^20?

1 個解答

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  • 2 個月前

     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)

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