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@Nooooo
August 2022
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ПОМОГИТЕ РЕШИТЬ ЛОГАРИФМ!!!!!
а) [tex] \frac{1}{4}^{1+0,5log_{ \frac{1}{2} } }^{14} [/tex]
б) [tex]25 ^{1-0,5log_{5} {11} } [/tex]
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Minsk00
A) (1/4)^(1+ 0,5log1/2(14))
(1/4)^(1+ 0,5log1/2(14)) = (1/2²)^(1+ 0,5log1/2(14)) = (2^(-2))^(1+ 0,5log1/2(14)) =
=(2^(-2))*(2^(-2))^(0,5log2^(-1)(14)) = (1/4)*(2^(-2))^(-0,5log2(14)) =
= (1/4)*(2^((-2)*(-0,5log2(14))) = (1/4)*(2^log2(14)) =(1/4)*14 =14/4 =7/2 =3,5
b) 25^(1 - 0,5log5(11)) = (5^(2))^(1 - 0,5log5(11)) = (5^(2))*(5^(2))^(-0,5log5(11)) =
=25*5^(2*(-0,5log5(11))) = 25*5^(-log5(11)) = 25*5^log5(1/11) = 25*(1/11) =25/11 =2 3/11
Ответ: a) 3,5; b) 25/11/
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Answers & Comments
(1/4)^(1+ 0,5log1/2(14)) = (1/2²)^(1+ 0,5log1/2(14)) = (2^(-2))^(1+ 0,5log1/2(14)) =
=(2^(-2))*(2^(-2))^(0,5log2^(-1)(14)) = (1/4)*(2^(-2))^(-0,5log2(14)) =
= (1/4)*(2^((-2)*(-0,5log2(14))) = (1/4)*(2^log2(14)) =(1/4)*14 =14/4 =7/2 =3,5
b) 25^(1 - 0,5log5(11)) = (5^(2))^(1 - 0,5log5(11)) = (5^(2))*(5^(2))^(-0,5log5(11)) =
=25*5^(2*(-0,5log5(11))) = 25*5^(-log5(11)) = 25*5^log5(1/11) = 25*(1/11) =25/11 =2 3/11
Ответ: a) 3,5; b) 25/11/