<s>[INST] I want to Find $3^{2x} + 7^{\frac{1}{x}}$ when $7^{x+1} = 21^{x}$ I figured that 
\begin{align*}7^{x+1} &= 21^{x}\\
7^{x+1} &= 3^{x} \times 7^{x} \\
7 &= 3^{x}
\end{align*}
 but I can't go further at the moment. 
 [/INST] You have : $7^{\frac{1}{x}} = 3$ and $3^{2x} = 7^2$. Thus $7^{\frac{1}{x}} + 3^{2x} = 52$
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