Помогите, пожалуйста (: 2 и 3 номер ♥️

Question

Помогите, пожалуйста (:
2 и 3 номер ♥️

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Comments

3) log(30)^2x вышло...

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4а)корень из b*кубичный корень из а

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3) 1

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не распишите?

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x≠1 * * * log₂ˣ =1/logₓ² ; log₃ˣ= 1/logₓ³ ; log₅ˣ = 1/logₓ⁵ * * *

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log30(x)=1/(logx(5)+logx(2)+logx(3))

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(log2(x)•log3(x)•log5(x))*(logx(5)+logx(2)+logx(3))=

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=log2(x)•log3(x)+log3(x)•log5(x)+log5(x)•log2(x)

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ответ 1

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благодарю!

Answer 1

2)log12(27)=a;log6(16)=?
log12(3^3)=3•log3(3)/log3(12)=3/(log3(3)+2log3(2))=3/(1+2log3(2))=a
3=a+2alog3(2)
log3(2)=(3-a)/2a
log6(16)=4log6(2)=4•log3(2)/log3(6)=
4log3(2)/(1+log3(2))=4(3-a)/2a:(1+(3-a)/2a)=
2(3-a)/a:(a+3)/2a=2(3-a)/a*2a/a+3=
4(3-a)/(3+a)
3)a)log30(x)=1/logx(30)=
1/(logx(5)+logx(2)+logx(3))
b)1/(log2(x)•log3(x)•log5(x))
*1/(logx(5)+logx(2)+logx(3))=
log2(x)•log3(x)+log3(x)•
log5(x)+log5(x)•log2(x)
c)(log2(x)•log3(x)+log3(x)•
log5(x)+log5(x)•log2(x)):
(log2(x)•log3(x)+log3(x)•
log5(x)+log5(x)•log2(x))
=1