Solve logarithmic equation $ 3^{\log_3^2x} + x^{\log_3x}=162$ Find $x$ from logarithmic equation $$ 3^{\log_3^2x} + x^{\log_3x}=162$$
I tried solving this, with basic logarithmic laws, changing base, etc., but with no result, then I went to wolframalpha and it says that its alternate form is:
$$2e^{\frac{\log^2x}{\log3}} = 162$$
But I don't know how it came to this result, can you help me guys?
 A: $3^{\log^2_3 x} = (3^{\log_3 x})^{\log_3 x} = x^{\log_3 x}$
A: The following is how WolframAlpha simplified it:
$$\begin{align}
162&=3^{\log_3^2x} + x^{\log_3x}
\\&=3^{\log_3x\log_3x} + x^{\log_3x}
\\&=\left(3^{\log_3x}\right)^{\log_3x} + x^{\log_3x}
\\&=x^{\log_3x} + x^{\log_3x}
\\&=2x^{\log_3x}
\\&=2\left(e^{\log x}\right)^{\log_3x}
\\&=2e^{\log x\log_3x}
\\&=2e^{\log x\frac{\log x}{\log 3}}
\\&=2e^{\frac{\log^2 x}{\log3}}
\end{align}$$
However, this is probably not the optimal process if you want to solve the problem; leaving things in terms of $\log_3x$ is helpful.
A: Since $a^{\log_a M}=M$ we have $x^{\log_3 x}+x^{\log_3 x}=2x^{\log_3 x}=162=2*3^4$.
Hence taking logarithms in base $3$ we have $(\log_3 x)^2=4$ so $(\log_3 x)=2$ (for positives) and $x=9$. Taking negatives we get $x=\frac 19$. There are two solutions.
A: All those $\log_3$'s make me want to think about what happens when $x$ is a power of $3$. 
$$
\begin{align}
3^{\log_3^23^k}+\left(3^k\right)^{\log_33^k}&=162\\
3^{k^2}+3^{k^2}&=162\\
3^{k^2}&=81\\
k^2&=4\\
k&=\pm2\\
x&=3^2\text{ or }3^{-2}
\end{align}$$
(Any positive $x$ can be written as $3^k$, and negative $x$ clearly can't be solutions since the LHS makes no sense for negative $x$.)
A: Introduce a new variable $y$ with $y=\log_3 x$.  
Then, we have $3^{y} = x \implies 3^{\log_3^2 x}=3^{y^2},\,\,\text{and}\,\,  x^{\log_3 x}=(3^{y})^{y}=3^{y^2}$.
Putting it together we have
$$3^{\log^2_3x}+x^{\log_3 x}=162\implies 2\times 3^{y^2}=162\implies y=\pm 2 \implies x=3^{\pm 2}$$  
