Question

solve for x 

$3^{\log^2_{3} x }  +  x^{\log_{3} x } = 162$

Answer 1

$\large 3^{\log_{3}^{2}x}+x^{\log_{3}x}=162$

=> $\large (3^{\log_{3}x})^{\log_{3}x}+x^{\log_{3}x}=162$

=>$\large (x)^{\log_{3}x}+x^{\log_{3}x}=162$                   [ $\large 3^{\log_{3}x}=x$]

=>$\large 2x^{\log_{3}x}=162$

=>$\large x^{\log_{3}x}=81$

taking log both the side with base 3,

=>$\large \log_{3}^{2}x=\log_{3}81$

=>$\large \log_{3}^{2}x=4$

=>$\large \log_{3}x=\pm 2$

so $\large x=9$ or $\large x=\frac{1}{9}$

Comments

I did not understand why you took (3^log3x) = x?

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@khushitshah

say ,

$\large z=3^{\log_{3}x}$

taking log on both side on base 3,

=> $\large \log_{3}z={\log_{3}x}$

=> $z=x$, provided $z$ and $x$ are positive real numbers i.e. $z,x>0$

so ,$3^{\log_{3}x}=x$

Edit Summary: added the statement "provided $z$ and $x$ are positive real numbers i.e. $z,x>0$" by ankitgupta.1729  

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@Kabir5454 Thanks got it!