0,4-log32 (x+3)=3log32 2
Ответы на вопрос
Ответил DimaPuchkov
1
ОДЗ: 
![\frac{2}{5} - \log_{32} (x+3) = \log_{32}2^3 \\ \\ \frac{2}{5} = \log_{32}8 + \log_{32} (x+3) \\ \\ \log_{32}(8(x+3)) = \frac{2}{5} \\ \\ 32^{\frac{2}{5}}=8x+24 \\ \\ 8x= \sqrt[5]{32^2} -24 \\ \\ 8x=\sqrt[5]{(2^5)^2}-24 \\ \\ 8x=4-24 \\ \\ 8x=-20; \ \ x =-\frac{20}{8}=-\frac{5}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B5%7D+-+%5Clog_%7B32%7D+%28x%2B3%29+%3D+%5Clog_%7B32%7D2%5E3+%5C%5C+%5C%5C+%5Cfrac%7B2%7D%7B5%7D++%3D+%5Clog_%7B32%7D8+%2B+%5Clog_%7B32%7D+%28x%2B3%29+%5C%5C+%5C%5C+%5Clog_%7B32%7D%288%28x%2B3%29%29++%3D+%5Cfrac%7B2%7D%7B5%7D++%5C%5C+%5C%5C+32%5E%7B%5Cfrac%7B2%7D%7B5%7D%7D%3D8x%2B24+%5C%5C+%5C%5C++8x%3D+%5Csqrt%5B5%5D%7B32%5E2%7D+-24+%5C%5C+%5C%5C++8x%3D%5Csqrt%5B5%5D%7B%282%5E5%29%5E2%7D-24+%5C%5C+%5C%5C+8x%3D4-24+%5C%5C+%5C%5C+8x%3D-20%3B+%5C+%5C+x+%3D-%5Cfrac%7B20%7D%7B8%7D%3D-%5Cfrac%7B5%7D%7B2%7D "\frac{2}{5} - \log_{32} (x+3) = \log_{32}2^3 \\ \\ \frac{2}{5} = \log_{32}8 + \log_{32} (x+3) \\ \\ \log_{32}(8(x+3)) = \frac{2}{5} \\ \\ 32^{\frac{2}{5}}=8x+24 \\ \\ 8x= \sqrt[5]{32^2} -24 \\ \\ 8x=\sqrt[5]{(2^5)^2}-24 \\ \\ 8x=4-24 \\ \\ 8x=-20; \ \ x =-\frac{20}{8}=-\frac{5}{2}")
Ответил natali3221
0
0.4- log32 (x+3)=3log32 2
0.4log32 (32)-log32(x+3)=log32( 2³) ОДЗ: x+3>0 x>-3
log32 (32^2\5) -log32 (x+3)=log32 (8) 32^2\5 =(2^5)^1\5)²=2²=4
log32 (4\(x+3)=log32 (8) основания одинаковы , логарифмы опускаем
4\(x+3)=8
8·(x+3)=4
8x+24=4
8x=4-24
8x=-20
x=-20:8
х=-2,5 -2.5>-3
Ответ: -2,5
0.4log32 (32)-log32(x+3)=log32( 2³) ОДЗ: x+3>0 x>-3
log32 (32^2\5) -log32 (x+3)=log32 (8) 32^2\5 =(2^5)^1\5)²=2²=4
log32 (4\(x+3)=log32 (8) основания одинаковы , логарифмы опускаем
4\(x+3)=8
8·(x+3)=4
8x+24=4
8x=4-24
8x=-20
x=-20:8
х=-2,5 -2.5>-3
Ответ: -2,5
Новые вопросы