Show that (7!)^1/7 < (8!)^1/8 without the use of a calculator.
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Show that (7!)^1/7 < (8!)^1/8 without the use of a calculator.

[From: ] [author: ] [Date: 12-08-27] [Hit: ]
7!so the property is proved-(7!)^8 = (7!)^7 * 7!= (7!= (8!......
How can I show that (7!)^1/7 is less than (8!)^1/8 without the use of a calculator? Please help.

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take ln of both sides

(1/7) (ln 2 + ln 3 + ln 4 + ln 5 + ln 6 + ln 7)

(1/8) (ln 2 + ln 3 + ln 4 + ln 5 + ln 6 + ln 7 + ln 8)

multiply both sides by 56

left side=
8 (ln 2 + ln 3 + ln 4 + ln 5 + ln 6 + ln 7)
[7 (ln 2 + ln 3 + ln 4 + ln 5 + ln 6 + ln 7)] + (ln 2 + ln 3 + ln 4 + ln 5 + ln 6 + ln 7)

right side =
7 (ln 2 + ln 3 + ln 4 + ln 5 + ln 6 + ln 7 + ln 8) =
= [7 (ln 2 + ln 3 + ln 4 + ln 5 + ln 6 + ln 7)] + 7ln 8

subtract the [ ] from both sides

left side = (ln 2 + ln 3 + ln 4 + ln 5 + ln 6 + ln 7) = ln 7!

right side = 7ln 8 = ln 8^7

7! is less than 8^7
so the property is proved

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(7!)^8 = (7!)^7 * 7!
< (7!)^7 * 8^7
= (7! * 8)^7
= (8!)^7

Taking both sides of the overall inequality to the 1/56 power gives (7!)^(1/7) < (8!)^(1/8).

Lord bless you today!
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