<s>[INST] Solve equation: $\frac{7}{\sqrt{343^{5x-1}}} = 2401^{-6.7}$ I need help to solve this equation, thanks in advance.
$$\frac{7}{\sqrt{343^{5x-1}}} = 2401^{-6.7}$$
 [/INST] $$\frac{7}{\sqrt{343^{5x-1}}} = 2401^{-6.7}$$
Since $343=7^3$ and $2401 = 7^4$, we can write
$$
\frac{7}{\sqrt{7^{3(5x-1)}}} = 7^{4\cdot(-6.7)}
$$
and then
$$
\frac{7}{\left(7^{3(5x-1)}\right)^{1/2}} = 7^{4\cdot(-6.7)}
$$
So
$$
7^{1 - (1/2)(3)(5x-1)} = 7^{4\cdot(-6.7)}.
$$
Hence
$$
1 - \frac12 \cdot3(5x-1) = 4\cdot(-6.7).
$$
etc.
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