Question 2447: (10pow 11) > (11 pow 10) and (20 pow 21) > (21 pow 20)please send the procedure to solve this problem
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! First of all , you put this question in wrong category. Secondly,
you should know how to type power correctly.
We can use logarithm to solve the inequalities.
To show 10^11 > 11^10
Let log be the common logarithm.
log 10^11 = 11,
log 11^10 = 10 log 11 = 10.41392685
Hence, log 10^11 > log 11^10
and so 10^11 > 11^10.
Similarly,
log 20^21 > 21 log 20 = 27.32
log 21^20 > 20 log 21 = 26.44
Hence, log 20^21 > log 20^21
and so 20^21 > 21^20.
Another way:
Consider 11^10/10^10 = (11/10)^10 = (1.1)^10 = (1.1^2)^5 = (1.21)^5
= (1.21)^2 * (1.21)^3 = 1.4641* 1.771561 < 1.5*2 = 3 < 10
Hence, 11^10 < 10* 10^10 = 10^11 and so 10^11 > 11^10.
Consider 21^20/20^20 = (21/20)^20 = (1.05^10)^2 < (1.1^10)^2
< 3^2 < 20
Hence, 21^20 < 20* 20^20 = 20^21 and so 20^21 > 21^20.
Kenny
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