SOLUTION: The expression (4b^−3 c^3 )^−5 (3b^−4 a^−2 )^−2 equals n(a^r)(b^s)(c^t) where n, the leading coefficient, is: and r, the exponent of a, is:

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: The expression (4b^−3 c^3 )^−5 (3b^−4 a^−2 )^−2 equals n(a^r)(b^s)(c^t) where n, the leading coefficient, is: and r, the exponent of a, is:       Log On


   

Question 275112: The expression (4b^−3 c^3 )^−5 (3b^−4 a^−2 )^−2 equals n(a^r)(b^s)(c^t)
where n, the leading coefficient, is:
and r, the exponent of a, is:
and s, the exponent of b, is:
and finally t, the exponent of c, is:
[NOTE: Your answers cannot be algebraic expressions.]
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
%284b%5E%28-3%29c%5E3%29%5E%28-5%29%2A%283b%5E%28-4%29a%5E%28-2%29%29%5E%28-2%29
We can start by using the property of exponents: %28a%2Ab%29%5Ep+=+a%5Ep%2Ab%5Ep to raise the two expressions in parentheses to the -5 and -2 powers respecitvely:

Using the property of exponents, %28a%5Ep%29%5Eq+=+a%5E%28p%2Aq%29, we get:

Rearranging the order and grouping of the factors, using the Commutative and Associative Properties of Multiplication, we get:
4%5E%28-5%29%2A3%5E%28-2%29%2Aa%5E4%2Ab%5E15%2Ab%5E8%2Ac%5E%28-15%29
Using the property of exponents, a%5Ep%2Aa%5Eq+=+a%5E%28p%2Bq%29 on the "b" factors we get:
4%5E%28-5%29%2A3%5E%28-2%29%2Aa%5E4%2Ab%5E%2815%2B8%29%2Ac%5E%28-15%29
which simplifies to:
4%5E%28-5%29%2A3%5E%28-2%29%2Aa%5E4%2Ab%5E23%2Ac%5E%28-15%29
This makes
n = 4%5E%28-5%29%2A3%5E%28-2%29+=+1%2F%284%5E5%2A3%5E2%29+=+1%2F%289%2A4%5E5%29 If this is an "algebaric expression", I'll leave it up to you to calculate 9%2A4%5E5.
r = 4
s = 23
t = -15