I want to Find $3^{2x} + 7^{\frac{1}{x}}$ when $7^{x+1} = 21^{x}$

Question

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.

Answer 1

Square both sides to reveal that

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

$x$ root both sides to reveal that

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

Thus,

$$3^{2x}+7^{1/x}=52$$

Answer 2

You have : $7^{\frac{1}{x}} = 3$ and $3^{2x} = 7^2$. Thus $7^{\frac{1}{x}} + 3^{2x} = 52$

Answer 3

Take the logarithm on both sides.