Respuesta :

It would be
2x3xnxn
N has a 1 in front of it it's just not shown
So
2x3x1nx1n then you'd at as much as possible
5x2n
[tex]2x^3*n*n=54[/tex] 

[tex]\ Use \ Product \ Rule[/tex]       [tex]x^ax^b=x^a^+^b[/tex] 

[tex]2x^3n^2=54[/tex] 

[tex]x^3n^2= \dfrac{54}{2} [/tex]      [tex]\ Divide \ both \ sides \ by \ 2[/tex] 

[tex]x^3n^2=27 [/tex] 

[tex]n^2= \dfrac{27}{x^3} [/tex]        [tex]\ Divide \ both \ sides \ by \ x^3[/tex] 

[tex]n=+ \sqrt{ \dfrac{27}{x^3}} [/tex] [tex] \ Take \ the \ square \ root \ of \ both \ sides[/tex] 

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[tex]If \ you \ mean \ 2x*3xnxn=54 . \ Lets \ Solve \ for \ n[/tex] 

[tex]2x*3xnxn=54[/tex] 

[tex]6x^3n^2=54 [/tex] 

[tex]x^3n^2= \dfrac{54}{6} [/tex]        [tex]\ Divide \ both \ sides \ by \ 6[/tex]

[tex]x^3n^2=9 [/tex] 

[tex]n^2= \dfrac{9}{x^3} [/tex]           [tex]\ Divide \ both \ sides \ by \ x^3[/tex]

[tex]n=+ \sqrt{ \dfrac{9}{x^3} } [/tex]  [tex] \ Take \ the \ square \ root \ of \ both \ sides[/tex]