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@Torlate
October 2021
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77 баллов!!! Нужна ваша помощь, завтра уже по этому материалу кр будет
Макс кол-во баллов!!!Помогите, очень прошу....
Нужно решить с номера 16.12 по 16.16 с объяснениями пожалуйста)))
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Tolianuch965
Примечание: степень пишу как ^ а логарифмы log(a)b - где а это основание а b число.
16.12 а) 2^(2+log(2)5) = 2^2+2^log(2)5 = 4+5 = 9
б) 5^(log(5)16 - 1) = 5^log(5)16 * (1/5) = 16/5
в) 3^(1+log(3)8) = 3^1 + 3^log(3)8 = 3+8 = 11
г) 8^(log(8)3 - 2)= 8^log(8)3 * (1/64) = 3/64
16.13
а) 2^(3log(2)4) = 4^3 = 64
б) (1/2)^(2log(1/2)7) = 7^2 = 49
в) 5^(2log(5)3) = 3^2 = 9
г) 0,3^(3log(0,3)6) = 6^3 = 216
16.14
а) 8^log(2)3 = 2^(3log(2)3) = 3^3 = 27
б) (1/9)^log(1/3)13 = 1/3^(2log(1/3)13) = 13^2 = 169
в) 25^log(5)3 = 5^(2log(5)3) = 3^2 = 9
г) (1/16)^log(1/2)5 = (1/2)^(4log(1/2)5) = 5^4 = 625
16.15
а) 36^((1/2)log(6)18) = 36^log(36)18 = 18
б) 64^((1/4)log(8)25) = 64^((1/2)log(64)25) = 25^(1/2) = 5
в) 121^((1/2)log(11)35) = 121^log(121)35 = 35
г) 25^((1/4)log(5)9) = 25^((1/2)log(25)9) = 9^(1/2) = 3
16.16
а) (1/4)^(1+(1/2)log(1/2)14) = (1/4) * (1/4)^log(1/4)14 = 1/4 * 14 = 14/4 = 7/2
б) 25^(1 - (1/2)log(5)11) = 25 * (1/(25^log(25)11)) = 25 * (1/11) = 25/11
в) (1/9)^(1+(1/2)log(1/3)18) = (1/9) * (1/9)^log(1/9)18 = (1/9) * 18 = 2
г) 49^(1 - (1/2)log(7)14) = 49 * (1/(49^log(49)14)) = 49 * (1/14) = 49/14 = 7/2
3 votes
Thanks 7
Tolianuch965
Если не затруднит поставьте "наилучший ответ")
Torlate
я хз как, оповещение само собой приходит и тогда отмечаю, как лучшее
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Answers & Comments
16.12 а) 2^(2+log(2)5) = 2^2+2^log(2)5 = 4+5 = 9
б) 5^(log(5)16 - 1) = 5^log(5)16 * (1/5) = 16/5
в) 3^(1+log(3)8) = 3^1 + 3^log(3)8 = 3+8 = 11
г) 8^(log(8)3 - 2)= 8^log(8)3 * (1/64) = 3/64
16.13
а) 2^(3log(2)4) = 4^3 = 64
б) (1/2)^(2log(1/2)7) = 7^2 = 49
в) 5^(2log(5)3) = 3^2 = 9
г) 0,3^(3log(0,3)6) = 6^3 = 216
16.14
а) 8^log(2)3 = 2^(3log(2)3) = 3^3 = 27
б) (1/9)^log(1/3)13 = 1/3^(2log(1/3)13) = 13^2 = 169
в) 25^log(5)3 = 5^(2log(5)3) = 3^2 = 9
г) (1/16)^log(1/2)5 = (1/2)^(4log(1/2)5) = 5^4 = 625
16.15
а) 36^((1/2)log(6)18) = 36^log(36)18 = 18
б) 64^((1/4)log(8)25) = 64^((1/2)log(64)25) = 25^(1/2) = 5
в) 121^((1/2)log(11)35) = 121^log(121)35 = 35
г) 25^((1/4)log(5)9) = 25^((1/2)log(25)9) = 9^(1/2) = 3
16.16
а) (1/4)^(1+(1/2)log(1/2)14) = (1/4) * (1/4)^log(1/4)14 = 1/4 * 14 = 14/4 = 7/2
б) 25^(1 - (1/2)log(5)11) = 25 * (1/(25^log(25)11)) = 25 * (1/11) = 25/11
в) (1/9)^(1+(1/2)log(1/3)18) = (1/9) * (1/9)^log(1/9)18 = (1/9) * 18 = 2
г) 49^(1 - (1/2)log(7)14) = 49 * (1/(49^log(49)14)) = 49 * (1/14) = 49/14 = 7/2