Steve S. answered • 03/13/14
Tutor
5
(3)
Tutoring in Precalculus, Trig, and Differential Calculus
Let’s do a similar problem:
E = 11 ( 2 a^3 b^(-5) )^(-7)
I spread the numbers apart for visibility, but multiplication is still assumed because there is no arithmetic operator symbol shown.
We know (x y^q )^p = x^p y^(pq), so:
E = 11 2^(-7) a^((3)(-7)) b^((-5)(-7))
E = 11 2^(-7) a^(–21) b^(35)
We know x^(-3) = 1/(x^3), so:
E = 11 (1/2^7) (1/a^21) b^35
E = 11 b^35 / (128 a^21)
E = 11 ( 2 a^3 b^(-5) )^(-7)
I spread the numbers apart for visibility, but multiplication is still assumed because there is no arithmetic operator symbol shown.
We know (x y^q )^p = x^p y^(pq), so:
E = 11 2^(-7) a^((3)(-7)) b^((-5)(-7))
E = 11 2^(-7) a^(–21) b^(35)
We know x^(-3) = 1/(x^3), so:
E = 11 (1/2^7) (1/a^21) b^35
E = 11 b^35 / (128 a^21)
