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Question Number 204647 by Engr_Jidda last updated on 24/Feb/24

Answered by Rasheed.Sindhi last updated on 24/Feb/24

((x/2))^((x/2)−1) =3^2   ⇐(x/2)=3 ∧ (x/2)−1=2⇒x=6

(x2)x21=32x2=3x21=2x=6

Commented by Engr_Jidda last updated on 24/Feb/24

thanks

thanks

Answered by Rasheed.Sindhi last updated on 24/Feb/24

((  ((x/2))^(x/2)   )/(x/2))=9=(3^3 /3)  (x/2)=3⇒x=6

(x2)x2x2=9=333x2=3x=6

Commented by Frix last updated on 24/Feb/24

There′s a 2^(nd)  solution we can only  approximste:  x≈.179044868

Theresa2ndsolutionwecanonlyapproximste:x.179044868

Commented by Frix last updated on 24/Feb/24

With x=2t we get  t^(t−1) =9  (t−1)ln t =2ln 3  Obviously t=3  0<t<1 ⇒ t−1<0∧ln t <0 ⇒ lhs>0  ⇒ there′s a solution in this interval

Withx=2twegettt1=9(t1)lnt=2ln3Obviouslyt=30<t<1t1<0lnt<0lhs>0theresasolutioninthisinterval

Answered by bello6646 last updated on 24/Feb/24

  solution     9 = 3^(3−1)      ((x/2))^((x/2)−1)  =  3^(3−1)         ⇒  (x/2)  =  3  ⇒ x  = 6

solution9=331(x2)x21=331x2=3x=6

Answered by lepuissantcedricjunior last updated on 26/Feb/24

((x/2))^((x/2)−1) =9=3^2   (x/2)ln((x/2))−ln((x/2))=2ln3=(6/2)ln((6/2))−ln((6/2)))  =>(x/2)=(6/2) par identif=> x=6

(x2)x21=9=32x2ln(x2)ln(x2)=2ln3=62ln(62)ln(62))=>x2=62paridentif=>x=6

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