Помогите 20 Балов!!!!!!
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Ответы на вопрос
Ответил spasibo3pajbrh
0
(1)
![log_{16}( log_{2}( log_{ sqrt[4]{2} }(x) ) ) = 0.5 \ log_{2}( log_{ sqrt[4]{2} }(x)) = sqrt{16} \ log_{ sqrt[4]{2} }(x) = {2}^{4} \ {4} log_{2}(x) = {2}^{4} \ log_{2}(x) = {2}^{2} \ x = {2}^{4} \ x = 16<br />](https://tex.z-dn.net/?f=+log_%7B16%7D%28+log_%7B2%7D%28+log_%7B+sqrt%5B4%5D%7B2%7D+%7D%28x%29+%29+%29+%3D+0.5+%5C+log_%7B2%7D%28+log_%7B+sqrt%5B4%5D%7B2%7D+%7D%28x%29%29+%3D+sqrt%7B16%7D+%5C+log_%7B+sqrt%5B4%5D%7B2%7D+%7D%28x%29+%3D+%7B2%7D%5E%7B4%7D+%5C+%7B4%7D+log_%7B2%7D%28x%29+%3D+%7B2%7D%5E%7B4%7D+%5C+log_%7B2%7D%28x%29+%3D+%7B2%7D%5E%7B2%7D+%5C+x+%3D+%7B2%7D%5E%7B4%7D+%5C+x+%3D+16%26lt%3Bbr+%2F%26gt%3B "log_{16}( log_{2}( log_{ sqrt[4]{2} }(x) ) ) = 0.5 \ log_{2}( log_{ sqrt[4]{2} }(x)) = sqrt{16} \ log_{ sqrt[4]{2} }(x) = {2}^{4} \ {4} log_{2}(x) = {2}^{4} \ log_{2}(x) = {2}^{2} \ x = {2}^{4} \ x = 16<br />")
(2)
![5 - 2 log_{7}(x) = 3 sqrt{ log_{7}(x) } \ y = sqrt{ log_{7}(x) } > 0\ 5 - 2{y}^{2} = 3y \ 2{y}^{2} + 3y - 5 = 0 \ 2(y - 1)(y + 2.5) = 0 \ y1 = 1 \ y2 = - 2.5 \](https://tex.z-dn.net/?f=5+-+2+log_%7B7%7D%28x%29+%3D+3+sqrt%7B+log_%7B7%7D%28x%29+%7D+%5C+y+%3D+sqrt%7B+log_%7B7%7D%28x%29+%7D+%26gt%3B+0%5C+5+-+2%7By%7D%5E%7B2%7D+%3D+3y+%5C+2%7By%7D%5E%7B2%7D+%2B+3y+-+5+%3D+0+%5C+2%28y+-+1%29%28y+%2B+2.5%29+%3D+0+%5C+y1+%3D+1+%5C+y2+%3D+-+2.5+%5C+ "5 - 2 log_{7}(x) = 3 sqrt{ log_{7}(x) } \ y = sqrt{ log_{7}(x) } > 0\ 5 - 2{y}^{2} = 3y \ 2{y}^{2} + 3y - 5 = 0 \ 2(y - 1)(y + 2.5) = 0 \ y1 = 1 \ y2 = - 2.5 \")
у2 нам не подходит
Возвращаемся к замене
(3)
![log_{10}(10 {x}^{2} ) log_{10}(x) = 1 \ (log_{10}(10) + 2 log_{10}(x) ) log_{10}(x) = 1 \ (1 + 2 log_{10}(x) )log_{10}(x) = 1 \ y = log_{10}(x) \ (1 + 2y)y = 1 \ 2 {y}^{2} + y - 1 = 0 \ 2(y + 1)(y - frac{1}{2} ) = 0 \ y_1= - 1 \ y_2 = frac{1}{2} \ \ log_{10}(x) = - 1 \ x = {10}^{ - 1} \ x_1 = 0.1 \ \ log_{10}(x) = 0.5 \ x = {10}^{0.5} \ x_ 2 = sqrt{10} \](https://tex.z-dn.net/?f=+log_%7B10%7D%2810+%7Bx%7D%5E%7B2%7D+%29+log_%7B10%7D%28x%29+%3D+1+%5C+%28log_%7B10%7D%2810%29+%2B+2+log_%7B10%7D%28x%29+%29+log_%7B10%7D%28x%29+%3D+1+%5C+%281+%2B+2+log_%7B10%7D%28x%29+%29log_%7B10%7D%28x%29+%3D+1+%5C+y+%3D+log_%7B10%7D%28x%29+%5C+%281+%2B+2y%29y+%3D+1+%5C+2+%7By%7D%5E%7B2%7D+%2B+y+-+1+%3D+0+%5C+2%28y+%2B+1%29%28y+-+frac%7B1%7D%7B2%7D+%29+%3D+0+%5C+y_1%3D+-+1+%5C+y_2+%3D+frac%7B1%7D%7B2%7D+%5C+%5C+log_%7B10%7D%28x%29+%3D+-+1+%5C+x+%3D+%7B10%7D%5E%7B+-+1%7D+%5C+x_1+%3D+0.1+%5C+%5C+log_%7B10%7D%28x%29+%3D+0.5+%5C+x+%3D+%7B10%7D%5E%7B0.5%7D+%5C+x_+2+%3D+sqrt%7B10%7D+%5C+ "log_{10}(10 {x}^{2} ) log_{10}(x) = 1 \ (log_{10}(10) + 2 log_{10}(x) ) log_{10}(x) = 1 \ (1 + 2 log_{10}(x) )log_{10}(x) = 1 \ y = log_{10}(x) \ (1 + 2y)y = 1 \ 2 {y}^{2} + y - 1 = 0 \ 2(y + 1)(y - frac{1}{2} ) = 0 \ y_1= - 1 \ y_2 = frac{1}{2} \ \ log_{10}(x) = - 1 \ x = {10}^{ - 1} \ x_1 = 0.1 \ \ log_{10}(x) = 0.5 \ x = {10}^{0.5} \ x_ 2 = sqrt{10} \")
(2)
у2 нам не подходит
Возвращаемся к замене
(3)
Ответил oganesbagoyan
0
4 - ая строка ?
Ответил spasibo3pajbrh
0
спасибо!;) исправлено!
Ответил oganesbagoyan
0
log₂ (x) =2² ⇒ x =2⁴ =16
Ответил spasibo3pajbrh
0
точно! спасибо! видимо, пора мне спать! доброй ночи!
Ответил oganesbagoyan
0
task/29887931
ответ : 1. → 16 ; 2 .→ 7 ; 3 . → 0. 1 ; √10
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Ответил spasibo3pajbrh
0
в первом задании: когда степень у основания логарифма ¼, то выносится за логарифм 4.
Ответил oganesbagoyan
0
logₓⁿ b =( logₓ b ) / n иначе (1/n)*(logₓ b) ... = (log₂ x) / (1/4) =4log₂ x
Ответил spasibo3pajbrh
0
да, да Вы правы!
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