I want to Find $3^{2x} + 7^{\frac{1}{x}}$ when $7^{x+1} = 21^{x}$ I figured that 
\begin{align*}7^{x+1} &amp;= 21^{x}\\
7^{x+1} &amp;= 3^{x} \times 7^{x} \\
7 &amp;= 3^{x}
\end{align*}
 but I can't go further at the moment.
Square both sides to reveal that

$$3^{2x}=49$$

$x$ root both sides to reveal that

$$7^{1/x}=3$$

Thus,

$$3^{2x}+7^{1/x}=52$$
Take the logarithm on both sides.