There should be 3 solutions right? By using De moivre's theorem we can find the roots of unity and have complex solutions and thus more solutions for X.
*_Absolutely_* ! 😁 We can say that *u^3* = 1 _Or_ => *(u^3)-1=0* Or =>(u - 1).(u^2+u+1)=0 , i.e. ,=> OF => *u* = 1 Or => *u{by quadratic formula} =[ -1± i√(3) ]/2* *Or* => *u ∈ {1,(w),(w^2)} , Where : (w) , (w^2) = imaginary Square roots of unity.(Complex-Conjugates of each other) Also , since *u= lg(x)* {Base 10 assumed} => Either x = 10 Or => x = antilog([ -1± i√3 ]/2])= 10^([ -1± i√3 ]/2]) ∴ *x ∈ { 10, 10^(w) , 10^(w^2) }* *(Ans.)* [Where : (w),(w^2) = imaginary Square roots of unity.(Complex conjugates of each other)]
Log(x)=y ,y should be a strictly positive number to even talk Abt solving this , keep this in mind √(y√y) =1 square both sides y√y =1 square again y³=1 , one real solutions , 2 complex ones , the only solution strictly positive is y=1 Thus logx=1 This x=10
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there will be 3 solutions right, 1 w and w^2
You could have saved a lot of time by squaring both sides twice (i.e. raising to the fourth power).
There should be 3 solutions right? By using De moivre's theorem we can find the roots of unity and have complex solutions and thus more solutions for X.
*_Absolutely_* ! 😁
We can say that *u^3* = 1 _Or_ => *(u^3)-1=0* Or =>(u - 1).(u^2+u+1)=0 , i.e. ,=> OF => *u* = 1 Or => *u{by quadratic formula} =[ -1± i√(3) ]/2* *Or* => *u ∈ {1,(w),(w^2)} , Where : (w) , (w^2) = imaginary Square roots of unity.(Complex-Conjugates of each other)
Also , since *u= lg(x)* {Base 10 assumed} => Either x = 10 Or => x = antilog([ -1± i√3 ]/2])= 10^([ -1± i√3 ]/2])
∴ *x ∈ { 10, 10^(w) , 10^(w^2) }* *(Ans.)* [Where : (w),(w^2) = imaginary Square roots of unity.(Complex conjugates of each other)]
Note that I edited , to replace (+/-) with ± , & "Belongs-to" with ∈
Log(x)=y ,y should be a strictly positive number to even talk Abt solving this , keep this in mind
√(y√y) =1 square both sides
y√y =1 square again
y³=1 , one real solutions , 2 complex ones
, the only solution strictly positive is y=1
Thus logx=1
This x=10