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?
Best Answer
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.