simplyfy expression division property of exponents

musicrocks429

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Mar 17, 2007
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how do you solve 4/ (2x)^-3
so far what i have done is multiple (2x)^-3 and ive gotten .125 which is -1/8
i dont know where to go from there or if im doing it right :?:
 
Hi musicrocks429! I admit, music does rock.

So far I so far what i have done is multiple[d] \(\displaystyle 2x^{\,-\,3}\) and 've gotten .125 which is \(\displaystyle \,-\,\frac{1}{8}\)


Since you have a variable in there and you have no idea what it is yet, it's impossible to get a value w/out a variable for it yet.

How do I solve:\(\displaystyle \L \;\frac{4}{2x^{\,-\,3}\) ?

Since we have multiplication on the bottom of the fraction, not addition, we can look at the problem like this:

\(\displaystyle \L \;\frac{4}{2}\,\cdot\,\frac{1}{x^{\,-\,3}\)

What's 4 divide by 2? 2

So now we have: \(\displaystyle \L \;\frac{2}{1}\,\cdot\,\frac{1}{x^{\,-\,3}}\,=\,\frac{2}{x^{\,-\,3}}\)

Now a basic rule of exponents says: \(\displaystyle x^{\,-\,3}\,=\,\frac{1}{x^3}\)

So substitute \(\displaystyle x^{\,-\,3}\) for \(\displaystyle \frac{1}{x^3}\): \(\displaystyle \L \;\frac{2}{\frac{1}{x^{3}}\)

So you simplified answer is: \(\displaystyle \L \;\frac{2}{1}\,\cdot\,\frac{x^3}{1}\,=\,2x^3\)

There you go! (I hope this is what you meant be "solve"; because it's impossible to solve for something without it being equal to anything).
 
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