hard prob: simplifying with negative exponents

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This problem has got my friends and I stumped.

. . .\(\displaystyle \L \mbox{25. }\, \left(\frac{3x^{\frac{-2}{3}}\, -\, 2x^{\frac{1}{5}}}{x^{\frac{-1}{3}}}\right)^{\frac{-1}{2}}\)

My work:

. . .\(\displaystyle \L \left(\frac{3x^{\frac{1}{3}}\, -\, 2x^{\frac{8}{15}}}{x^{\frac{2}{3}}}\right)^{\frac{-1}{2}}\)

. . .\(\displaystyle \L \frac{\left(x^{\frac{2}{3}}\right)^{\frac{1}{2}}}{\left(3x^{\frac{1}{3}}\, -\, 2x^{\frac{8}{15}}\right)^{\frac{1}{2}}}\, =\, \frac{x^{\frac{2}{6}}}{3x^{\frac{1}{6}}\, -\, 2x^{\frac{8}{36}}}\)

. . .\(\displaystyle \L \frac{x^{\frac{1}{3}}}{3x^{\frac{1}{6}}\, -\, 2x^{\frac{4}{15}}}\)

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Edited by stapel -- Reason for edit: removing horizontal scroll
 
You can cancel only factors, not terms. To cancel off, you would have to factor something out of the numerator first; you can't just hack off parts of each term.

You can take a power through onto the factors of a product, but not onto the terms in a sum.

Try using the regular rules for fractions. Start by getting a common denominator, and combining. One way to go might be:

. . .3x^(-2/3) - 2x^(1/5)

. . . . .= 3/(x^(2/3) - 2x^(1/5)

. . . . .= 3x^(1/3)/x - 2x^(6/5)/x

. . . . .= [3x^(1/3) - 2x^(6/5)] / x

. . .x^(-1/3) = 1/x^(1/3) = x^(2/3) / x

And so forth.

Eliz.
 
Since n^(-1/2) means 1 / sqrt(n), forget about the ^(-1/2) : get back to it at end;

[3x^(-2/3) - 2x^(1/5)] / x^(-1/3)

= [3 / x^(2/3) - 2x^(1/5)] / [1 / x^(1/3)]

= {[3 - 2x^(1/5) x^(1/3)] / x^(2/3)} / [1 / x^(1/3)]

= [x^(1/3)][3 - 2x^(13/15)] / [x^(2/3)]

= [3x^(1/3) - 2x^(6/5)] / [x^(2/3)] : 1/3 + 13/15 = 6/5

So, with the ^(-1/2):

1 / sqrt{ [3x^(1/3) - 2x^(6/5)] / [x^(2/3)] }
or
1 / {[3x^(1/3) - 2x^(6/5)] / [x^(2/3)]}^(1/2)

If you can't follow that, see your teacher about getting some additional help.
 
Hello, AirForceOne!

You made a few illegal moves in there . . .


\(\displaystyle \L25)\;\;\left(\frac{3x^{-\frac{2}{3}}\, -\, 2x^{\frac{1}{5}}}{x^{-\frac{1}{3}}}\right)^{-\frac{1}{2}}\)

Inside the parentheses, multiply top and bottom by \(\displaystyle x^{\frac{2}{3}\)

. . . . \(\displaystyle \L\:\frac{x^{\frac{2}{3}}}{x^{\frac{2}{3}}}\,\cdot\,\frac{3x^{-\frac{2}{3}}\,-\,2x^{\frac{1}{5}}}{x^{-\frac{1}{3}}} \;= \;\frac{3\,-\,2x^{\frac{13}{15}}}{x^{\frac{1}{3}}}\)


Then we have: \(\displaystyle \L\:\left(\frac{3\,-\,2x^{\frac{13}{15}}}{x^{\frac{1}{3}}}\right)^{-\frac{1}{2}} \;= \;\left(\frac{x^{\frac{1}{3}}}{3\,-\,2x^{\frac{13}{15}}}\right)^{\frac{1}{2}} \;=\;\frac{\left(x^{\frac{1}{3}}\right)^{\frac{1}{2}}}{\left(3\,-\,2x^{\frac{13}{15}}\right)^{\frac{1}{2}}}\)

Answer: \(\displaystyle \L\:\frac{x^{\frac{1}{6}}}{\sqrt{3\,-\,2x^{\frac{13}{15}}}}\)

 
Sorry for the late reply.

Thanks to all, but Soraban, don't you have to remove the radical from the bottom of the fraction in your final answer?

Thanks guys!
 
AirForceOne said:
...don't you have to remove the radical from the bottom of the fraction in your final answer?
Um... It kinda depends. When you're in calculus, it might not matter. After calculus, it almost certainly doesn't matter. Generally in algebra, it does.

So make the appropriate adjustment, if necessary for your class. But it doesn't change the answer mathematically.

Eliz.
 
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