Решите уравнение, подробно
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Ответы на вопрос
Ответил Artem112
0
ОДЗ:
![3x+1>0
\
3x>-1
\
x>- frac{1}{3}](https://tex.z-dn.net/?f=3x%2B1%26gt%3B0%0A%5C%0A3x%26gt%3B-1%0A%5C%0Ax%26gt%3B-+frac%7B1%7D%7B3%7D+ "3x+1>0
\
3x>-1
\
x>- frac{1}{3}")
Решаем:
![log_2^2(3x+1)-3log_ frac{1}{2} frac{4}{3x+1} =( frac{2}{7} )^{log_ frac{2}{7}1,5+log_ frac{2}{7}4 }
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log_2^2(3x+1)-3log_{2^{-1}} frac{4}{3x+1} =( frac{2}{7} )^{log_ frac{2}{7}(1,5cdot4) }
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log_2^2(3x+1)+3log_{2} frac{4}{3x+1} =( frac{2}{7} )^{log_ frac{2}{7}6 }
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log_2^2(3x+1)+3(log_{2}4-log_{2}(3x+1)) =6
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log_2^2(3x+1)+3(2-log_{2}(3x+1)) =6
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log_2^2(3x+1)+6-3log_{2}(3x+1)=6
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log_2^2(3x+1)-3log_{2}(3x+1)=0
\
log_2(3x+1)(log_{2}(3x+1)-3)=0](https://tex.z-dn.net/?f=log_2%5E2%283x%2B1%29-3log_+frac%7B1%7D%7B2%7D++frac%7B4%7D%7B3x%2B1%7D+%3D%28+frac%7B2%7D%7B7%7D+%29%5E%7Blog_+frac%7B2%7D%7B7%7D1%2C5%2Blog_+frac%7B2%7D%7B7%7D4+%7D%0A%5C%0Alog_2%5E2%283x%2B1%29-3log_%7B2%5E%7B-1%7D%7D++frac%7B4%7D%7B3x%2B1%7D+%3D%28+frac%7B2%7D%7B7%7D+%29%5E%7Blog_+frac%7B2%7D%7B7%7D%281%2C5cdot4%29+%7D%0A%5C%0Alog_2%5E2%283x%2B1%29%2B3log_%7B2%7D++frac%7B4%7D%7B3x%2B1%7D+%3D%28+frac%7B2%7D%7B7%7D+%29%5E%7Blog_+frac%7B2%7D%7B7%7D6+%7D%0A%5C%0Alog_2%5E2%283x%2B1%29%2B3%28log_%7B2%7D4-log_%7B2%7D%283x%2B1%29%29+%3D6%0A%5C%0Alog_2%5E2%283x%2B1%29%2B3%282-log_%7B2%7D%283x%2B1%29%29+%3D6%0A%5C%0Alog_2%5E2%283x%2B1%29%2B6-3log_%7B2%7D%283x%2B1%29%3D6%0A%5C%0Alog_2%5E2%283x%2B1%29-3log_%7B2%7D%283x%2B1%29%3D0%0A%5C%0Alog_2%283x%2B1%29%28log_%7B2%7D%283x%2B1%29-3%29%3D0 "log_2^2(3x+1)-3log_ frac{1}{2} frac{4}{3x+1} =( frac{2}{7} )^{log_ frac{2}{7}1,5+log_ frac{2}{7}4 }
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log_2^2(3x+1)-3log_{2^{-1}} frac{4}{3x+1} =( frac{2}{7} )^{log_ frac{2}{7}(1,5cdot4) }
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log_2^2(3x+1)+3log_{2} frac{4}{3x+1} =( frac{2}{7} )^{log_ frac{2}{7}6 }
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log_2^2(3x+1)+3(log_{2}4-log_{2}(3x+1)) =6
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log_2^2(3x+1)+3(2-log_{2}(3x+1)) =6
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log_2^2(3x+1)+6-3log_{2}(3x+1)=6
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log_2^2(3x+1)-3log_{2}(3x+1)=0
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log_2(3x+1)(log_{2}(3x+1)-3)=0")
1)
![log_2(3x+1)=0
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3x+1=2^0
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3x+1=1
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3x=0
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x_1=0](https://tex.z-dn.net/?f=log_2%283x%2B1%29%3D0%0A%5C%0A3x%2B1%3D2%5E0%0A%5C%0A3x%2B1%3D1%0A%5C%0A3x%3D0%0A%5C%0Ax_1%3D0 "log_2(3x+1)=0
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3x+1=2^0
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3x+1=1
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3x=0
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x_1=0")
2)
![log_{2}(3x+1)-3=0
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log_{2}(3x+1)=3
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3x+1=2^3
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3x+1=8
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3x=7
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x_2= frac{7}{3}](https://tex.z-dn.net/?f=log_%7B2%7D%283x%2B1%29-3%3D0%0A%5C%0Alog_%7B2%7D%283x%2B1%29%3D3%0A%5C%0A3x%2B1%3D2%5E3%0A%5C%0A3x%2B1%3D8%0A%5C%0A3x%3D7%0A%5C%0Ax_2%3D+frac%7B7%7D%7B3%7D+ "log_{2}(3x+1)-3=0
\
log_{2}(3x+1)=3
\
3x+1=2^3
\
3x+1=8
\
3x=7
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x_2= frac{7}{3}")
Оба корня попадают в ОДЗ.
Ответ: 0 и 7/3
Решаем:
1)
2)
Оба корня попадают в ОДЗ.
Ответ: 0 и 7/3
Ответил gorynich
0
Спасибо!
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