Question

Which is the simplified form of the expression ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript negative 3 Baseline times ((2 Superscript negative 3 Baseline) (3 squared)) squared?

Answer 1

Answer:

$\dfrac{1}{6561}$

Step-by-step explanation:

Given the expression $[(2^{-2})(3^{4})]^{-3} * [(2^{-3})(3^2)]^{2}$, Using the laws of indices to simplify the expression. The following laws will be applicable;

$a^m*a^n = a^{m+n}\\(a^m)^n = a^{mn}$

$a^{-m} = 1/a^m$

Given $[(2^{-2})(3^{4})]^{-3} * [(2^{-3})(3^2)]^{2}$

open the parenthesis

$= (2^{-2})^{-3}(3^{4})^{-3}* (2^{-3})^2(3^2)^2\\\\= 2^{-2*-3}* 3^{4*-3} * 2^{-3*2} * 3^{2*2}\\\\= 2^6 * 3^{-12} * 2^{-6} * 3^4\\\\collecting \ like \ terms\\\\= 2^6 * 2^{-6} * 3^{-12} * 3^4\\\\= 2^{6-6} * 3^{-12+4}\\\\= 2^0 * 3^{-8}\\\\= 1 * \frac{1}{3^8}\\ \\= \frac{1}{6561}$