SOLUTION: I don't understand how to simplify this question: (16 raised to the negative 1 power, "x" raised to the negative 4th power, "y" raised to the possitive 12th power) This whole pr

Algebra ->  Exponents-negative-and-fractional -> SOLUTION: I don't understand how to simplify this question: (16 raised to the negative 1 power, "x" raised to the negative 4th power, "y" raised to the possitive 12th power) This whole pr      Log On


   

Question 95098: I don't understand how to simplify this question:
(16 raised to the negative 1 power, "x" raised to the negative 4th power, "y" raised to the possitive 12th power) This whole problem is in parenthesis. Outside of the parenthesis, the entire problem is raisd to negative 3/4.

I hope this makes sense.
So I think it should be written as:
(16^-1 x^-4 y^12)^-3/4


Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given:
.
%28%2816%5E%28-1%29%29%28+x%5E%28-4%29+%29%28y%5E12%29%29%5E%28-3%2F4%29+
.
By the power rule of exponents you multiply each of the exponents inside the parentheses
by the -3%2F4.
.
When you do that the problem becomes:
.

.
Note ... for some reason the 3 in the numerator of each multiplication gets clipped off.
Hope this doesn't confuse you.
.
Do the multiplication by -3%2F4 in each of the exponents and you get:
.
%28%2816%5E%283%2F4%29%29%2A%28+x%5E%283%29%29%2A%28y%5E%28-9%29%29%29
.
At this point, let's just focus on 16%5E%283%2F4%29. We can work the power rule in reverse and
get two different forms of this as follows:
.
Either %2816%5E%281%2F4%29%29%5E3 or %2816%5E3%29%5E%281%2F4%29%29
.
Applying the power rule to either of these gets you back to 16%5E%283%2F4%29.
.
So to evaluate this term we can either find the fourth root of 16 and cube it, or we can
cube 16 and take the fourth root of that answer. In this case it's easier to find the
fourth root of 16 which is 2 and cube that to get 2%5E3+=+8 as the answer. So now we
can replace 16%5E%283%2F4%29 with 8 and the problem is then down to:
.
8x%5E3y%5E%28-9%29
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But if a term in the numerator has a minus exponent, you can put that same term in the
denominator with a positive exponent. So we can change y%5E%28-9%29 by putting it in the
denominator as y%5E9 and the answer to this problem becomes:
.
8x%5E3%2Fy%5E9
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Notice that in taking the fourth root of x and y, to get real answers we must limit
both of these variables to positive values. Therefore, we need to make the answer have
positive values of x and y which we can do through absolute values. Therefore, the answer is:
.
8%2A%28abs%28x%29%29%5E3%2F%28abs%28y%29%29%5E9
.
Hope this helps you to understand the problem.