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ДинA
@ДинA
August 2021
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1)log x по основанию(1/5) [tex] \leq [/tex] log(1/8) по основанию(1/5)
2)ln x БОЛЬШЕ ln 0,5
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Незнaйкa
1)Log1/5 x<=Log1/5 1/8
X>=1/8
[1/8;+бесконечности)
2)Ln x>Ln 0,5
X>0,5
(0,5;+бесконечности)
1 votes
Thanks 2
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Answers & Comments
X>=1/8
[1/8;+бесконечности)
2)Ln x>Ln 0,5
X>0,5
(0,5;+бесконечности)