Solve equation: $\frac{7}{\sqrt{343^{5x-1}}} = 2401^{-6.7}$ I need help to solve this equation, thanks in advance.
$$\frac{7}{\sqrt{343^{5x-1}}} = 2401^{-6.7}$$
 A: $$\frac{7}{\sqrt{343^{5x-1}}} = 2401^{-6.7}$$
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.
A: Note that $343 =7^3$ and $2401=7^4$. Applying logarithms gets rid of all the nuisance:
$$\begin{align*}
\frac{7}{\sqrt{343^{5x-1}}} &= 2401^{-6.7}\\
\log(7) - \frac{5x-1}{2}\log(343) &= -6.7\log(2401)\\
\log 7 - \frac{3(5x-1)}{2}\log 7 &= -26.8\log(7).
\end{align*}$$
At this point, it should be clear how to finish it off easily.
A: To avoid getting confused with numbers, write the equation as $$ \frac {a}{\sqrt{b^y}} = c \mbox,$$ solve for $y$, and then for $x$, using $y=5x-1$.
A: Do you know how to solve $\dfrac{7}{2401^{-6.7}} = \sqrt{343^{5x-1}}$ ? Further, do you know that, for example, $\dfrac{1}{4^{-2}} = 4^2 = 16$?
If you know these, then you are set. Perhaps I will also mention that that square root just gets in the say, so you should get rid of it. (do you know how?)
