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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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Where are you stuck? –  lhf Sep 5 '11 at 16:16
    
You know that $\sqrt{x}=x^{1/2}$ and that $\frac1{x}=x^{-1}$, don't you? Then you can use the properties of exponents... –  J. M. Sep 5 '11 at 16:17
    
i dont know how to simplify the right side of the equation. that power confuse me, because of that decimal power. i think that must be written as fraction, but still got no idea how to get some simpler expression. –  jbennet Sep 5 '11 at 16:21
    
@J.M. - i know that rule, but this is some kind of unusual exam, and the numbers on the right side are nebulous. there must be some trick, but i cant see it. –  jbennet Sep 5 '11 at 16:24
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Another way of getting those pesky exponents down on the level would be to take logarithms, but you'll do better with questions like this if you see where J.M.'s hint leaves you, and if you are expecting similar questions from the same source in future it would be well worth thinking about how you might check for something similar in a future question. –  Mark Bennet Sep 5 '11 at 16:38

4 Answers 4

up vote 4 down vote accepted

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

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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I don't like using logarithms either (unless the problem presses me to use them). I will upvote in a few hours... –  J. M. Sep 5 '11 at 17:19
    
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.) –  Michael Hardy Sep 7 '11 at 14:12

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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thanks a lot, that is what i neeed. x = 3.906666666 –  jbennet Sep 5 '11 at 16:55
    
Would have posted the same, very good approach! –  Gerben Sep 5 '11 at 17:26
    
@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. –  Arturo Magidin Sep 5 '11 at 19:30

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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yes i know that, but it's a harder way than method using logarithms due to big numbers on the one side of the equation, so i think that a professor want to solve it using logarithms. –  jbennet Sep 5 '11 at 17:10
    
@jbennet Oh, well I wish you had mentioned that you wanted to use logs. That's not so bad to write up either –  mixedmath Sep 5 '11 at 17:14
    
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. –  jbennet Sep 5 '11 at 17:18

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