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diman3104
@diman3104
September 2021
1
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Прошу,помогите!
Срочно!
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tolya91
1
а),,,=-11
б)...=log₅(5⁻⁴)=-4
в)...=log(₂^(1/2))(2³)=3/(1/2)=6
2
25·5^(log₅4) - 3·3^(log₃5)=25·4-3·5=85
3
а)lgx=3
10³=x
x=1000
б)ОДЗ: х≠1 х>0
x³=125
x=5
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diman3104
спасибо огромное
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Answers & Comments
а),,,=-11
б)...=log₅(5⁻⁴)=-4
в)...=log(₂^(1/2))(2³)=3/(1/2)=6
2
25·5^(log₅4) - 3·3^(log₃5)=25·4-3·5=85
3
а)lgx=3
10³=x
x=1000
б)ОДЗ: х≠1 х>0
x³=125
x=5