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aleksen
@aleksen
August 2022
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Хильмилли
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А) log8(64*2^(1/4)=log2(2^6*2^(1/4))/log2(2^3)=log2(2^(25/4)/log2(2^3)=
=(25/4)/3=25/12, так как 2^6*2^(1/4)=2^(25/4)
б) 25/(1-log5(10)=25^1/(25^(log5(10))=25/(5^(2*log5(10)=25/(5^(log5(10))^2=
=25/(10^2)=25/100=0.25
в) log5(x+3)=2-log5(2x+1)⇒ log5(x+3)=log5(25)-log5(2x+1)
log5(x+3)=log5(25/(2x+1))⇒x+3=25/(2x+1)⇒(x+3)*(2x+1)=25
2x^2+7x+3-25=0; 2x^2+7x-22=0
x1=(-7+√49+176)/4=(-7+15)/4=2
x2=(-7-√49+176)/4=(-7-15)/4=-11/2
г)(log3(x))^2-log3(3x)=1⇒(log3(x))^2-log3(3)-log3(x)=1
log3(x)=t⇒ t^2-t-2=0⇒t1+t2=1; t1*t2=-2; t1=2; t2=-1
log3(x)=2⇒x1=9
log3(x)=-1⇒x2=1/3
3 votes
Thanks 7
hote
в третьем примере ОДЗ Х>-0.5 Значит корень х=-11/2 не подходит
hote
Решение верное с учетом комментария
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Answers & Comments
Verified answer
А) log8(64*2^(1/4)=log2(2^6*2^(1/4))/log2(2^3)=log2(2^(25/4)/log2(2^3)==(25/4)/3=25/12, так как 2^6*2^(1/4)=2^(25/4)
б) 25/(1-log5(10)=25^1/(25^(log5(10))=25/(5^(2*log5(10)=25/(5^(log5(10))^2=
=25/(10^2)=25/100=0.25
в) log5(x+3)=2-log5(2x+1)⇒ log5(x+3)=log5(25)-log5(2x+1)
log5(x+3)=log5(25/(2x+1))⇒x+3=25/(2x+1)⇒(x+3)*(2x+1)=25
2x^2+7x+3-25=0; 2x^2+7x-22=0
x1=(-7+√49+176)/4=(-7+15)/4=2
x2=(-7-√49+176)/4=(-7-15)/4=-11/2
г)(log3(x))^2-log3(3x)=1⇒(log3(x))^2-log3(3)-log3(x)=1
log3(x)=t⇒ t^2-t-2=0⇒t1+t2=1; t1*t2=-2; t1=2; t2=-1
log3(x)=2⇒x1=9
log3(x)=-1⇒x2=1/3