\documentclass{article}

\usepackage{amsmath, amsthm, amssymb, amsfonts, mathtools}

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\begin{document}
% (2)
Substitute \(t = \log_{e}{x} = \ln{x}\).
\begin{align*}
(0.2^{(6t - 3)})^{1/t} &> (0.008^{2t - 1})^{1/3} \\ 
0.2^{(6t - 3)/t} &> (0.2^{3})^{(2t - 1)/3} \\ 
0.2^{(6t - 3)/t} &> 0.2^{3 \cdot [(2t - 1)/3]} \\ 
\frac{6t - 3}{t} &> 3 \cdot \frac{2t - 1}{3} \\ 
t \cdot \frac{6t - 3}{t} &> (2t - 1) \cdot t \\ 
6t - 3 &> 2t^2 - t \\ 
-2t^{2} + 7t - 3 &> 0 \\
2t^2 - 7t + 3 &< 0 \\
(2t - 1)(t - 3) &< 0 \\
\end{align*}

When \(t < \frac{1}{2}\) or \(t > 3\), the product of the factored terms is positive, so the inequality doesn't hold.\\
When \(\frac{1}{2} < t < 3\), the product of the factored terms is negative, so the inequality holds.\\
Substituting the original value of \(t\), we have that

\begin{equation*}
    \boxed{\frac{1}{2} < \ln{x} < 3}
\end{equation*}

%(3)
\begin{align*}
    3^{\ln{x} + \ln{x^0} + \ln{x^0} + \cdots + \ln{x^0}} &= 27x^{30} \\
    3^{\ln{x} + \ln{1} + \ln{1} + \cdots + \ln{1}} &= 27x^{30} \\
    3^{\ln{x} + 0 + 0 + \cdots + 0} &= 27x^{30} \\
    3^{\ln{x}} &= 27x^{30} \\
    \log_3{3^{\ln{x}}} &= \log_3{27x^{30}} \\
    \ln{x} &= \log_3{3^{3}} + \log_3{x^{30}} \\
    \ln{x} &= 3 + \log_3{x^{30}} \\
    \ln{x} &= 3 + \log_3{x^{30}} \\
    \ln{x} &= 3 + \frac{\ln{x^{30}}}{\ln{3}} \\
    \ln{x} &= 3 + \frac{30\ln{x}}{\ln{3}} \\
    \ln{3} \cdot \ln{x} &= 3\ln{3} + 30\ln{x} \\
    0 &= 3\ln{3} + 30\ln{x} - \ln{3} \cdot \ln{x} \\
    (30 - \ln{3})\ln{x} &= -3\ln{3} \\
    \ln{x} &= \frac{-3\ln{3}}{30 - \ln{3}}\\
    e^{\ln{x}} &= e^{\frac{-3\ln{3}}{30 - \ln{3}}}\\
    \Aboxed{x &= e^{\frac{-3\ln{3}}{30 - \ln{3}}}}
\end{align*}

\end{document}