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I need help to solve this equation, thanks in advance.

$\frac{7}{\sqrt{343^{5x-1}}} = 2401^{-6.7}$

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    @jbennet: I think you have models below from Michael Hardy ("without logarithms" - although equating exponents amounts to the same thing - the insight is that everything can be written as a power of 7) and Arturo Magidin (with logarithms - which gets you started without noticing that base 7 is a natural base for this question, so the logarithm is natural, or to base 10). Arturo's solution notices the status of 7 later, but using tables or a calculator, his method could be finished off without. It's frequently useful to notice special features (7) which make for an easier solution.2011-09-06

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$\frac{7}{\sqrt{343^{5x-1}}} = 2401^{-6.7}$

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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    In a sense this does use logarithms, in the last step, but it doesn't use algebraic properties of logarithms, beyond the trivial one that is essentially the last step above. The reason I posted this after others had been posted was actually to show that this could be done in a way that could be understood by people not familiar with those properties. The last step does tacitly assume the reader knows that exponential functions are one-to-one, i.e. $7^a$ cannot be the same as $7^b$ unless $a=b$. (That changes once you allow imaginary numbers.)2011-09-07
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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.

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    @jbennet: It would be much better if you express $x$ as an irreducible fraction, rather than as a decimal approximation to its actual value. At least, that's what I always tell my students.2011-09-05
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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 $y=5x-1$.

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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?)

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    noo, it's ok. in fact, i also thought that must be solved as normal equation, but logarithmic method is faster and easier. this task is worth 8% of exam, so i knew there is some relative easy way to do it.2011-09-05