SOLUTION: My goal is to simplify the expression with rational exponents: (25/49) to the -3/2 power: My answer is opposite of what the textbook provides. The way I solved it is as follo

Algebra ->  Square-cubic-other-roots -> SOLUTION: My goal is to simplify the expression with rational exponents: (25/49) to the -3/2 power: My answer is opposite of what the textbook provides. The way I solved it is as follo      Log On


   

Question 71665: My goal is to simplify the expression with rational exponents:
(25/49) to the -3/2 power: My answer is opposite of what the textbook provides. The way I solved it is as follows:
(25) -3/2 = The cubic root of 25 is (5) and -3 is my exponent, which is -125
(49) The cubic root of 49 is (7) and -3 is my exponent, which is -343
The answer I am getting is 125/343 because I cancelled out the negatives. The textbook is giving the answer of 343/125. I am not sure where my mistake is.

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
(25/49) to the -3/2 power
.
You have made a few mistakes in your description of your work. First you said that the
cube root of 25 is 5. Actually the square root of 25 is 5. The same goes for 49. You said
the cube root of 49 is 7. And actually the square root of 49 is 7. However, it seems that
the only mistake here was in saying that the square root was the cube root. I think where
you mistake occurred was in how you handled the negative exponents. The way to do negative
exponents is to use the form:
.
x%5E%28-a%29+=+1%2Fx%5E%28%2Ba%29
.
You can treat this as a definition of negative exponents.
.
In this problem 5%5E%28-3%29+=+1%2F5%5E%28%2B3%29+=+1%2F125
.
A similar thing takes place with the 49. The square root of 49 is 7 and then when you
raise 7 to the -3 power it tracks as follows: 7%5E%28-3%29+=+1%2F7%5E%28%2B3%29+=+1%2F343
and the
answer to you problem is obtained by dividing 1%2F125 by 1%2F343. Recall that when
you divide by a fraction, you invert the divisor and multiply the inverted form by the
number you are dividing. So in this case you invert 1%2F343 to get 343%2F1
and then you multiply that times 1%2F125. The result is:
.
%281%2F125%29%2A%28343%2F1%29+=+343%2F125
.
This answer agrees with the book. It appears to me that you understood that the exponent
of -3%2F2 was applied to the entire fraction 25%2F49 but that you could then apply
that exponent to the numerator and then apply it again to the denominator.
.
It also appeared to me that you understood that %2825%29%5E%28-3%2F2%29 could be written
either as
.
%28%2825%29%5E%28-3%29%29%5E%281%2F2%29
.
or as
.
%28%2825%29%5E%281%2F2%29%29%5E%28-3%29
.
and in both of these two formats the minus sign could also be put on the 1%2F2
leaving the 3 with a + sign. So there are four possible ways to work this problem and
each of them works if you do them correctly.
.
You chose for both the 25 and the 49 to use the format:
.
%28%2825%29%5E%281%2F2%29%29%5E%28-3%29 and %28%2849%29%5E%281%2F2%29%29%5E%28-3%29
.
but just canceled the minus signs on the 3's instead of using the method of eliminating
negative exponents by making a fraction with 1 as the numerator and the number now
raised to a positive exponent in the denominator.
.
Hope this helps a little to correct how you handle negative exponents. You were on the
right track with everything else.