Question

Help me? idk the answer :P

Answer 1

`\bf \left.\qquad \qquad \right.\textit{negative exponents}\\\\ a^{-{ n}} \implies \cfrac{1}{a^( n)} \qquad \qquad \cfrac{1}{a^( n)}\implies a^{-{ n}} \qquad \qquad a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}\\\\ -------------------------------\\\\ \left( 2^8\cdot 3^(-5)\cdot 6^0 \right)^(-2)\left( \cfrac{3^(-2)}{2^3} \right)^4\cdot 2^(28)\impliedby \textit{let's do the first group} \\\\ -------------------------------\\\\`


`\bf \left( 2^8\cdot \cfrac{1}{3^5}\cdot 1 \right)^(-2)\implies \left( \cfrac{2^8}{3^5} \right)^(-2)\implies \left( \cfrac{3^5}{2^8} \right)^(2)\implies \cfrac{3^(2\cdot 5)}{2^(2\cdot 8)}\implies \boxed{\cfrac{3^(10)}{2^(16)}}\\\\ -------------------------------\\\\ \textit{now the second group}\qquad \left( \cfrac{3^(-2)}{2^3} \right)^4\implies \left( \cfrac{(1)/(3^2)}{2^3} \right)^4\implies \left( \cfrac{1}{2^3\cdot 3^2} \right)^4`


`\bf \cfrac{1^4}{2^(4\cdot 3)\cdot 3^(4\cdot 2)}\implies \boxed{\cfrac{1}{2^(12)\cdot 3^8}}\\\\ -------------------------------\\\\ \textit{so we end up with\qquad }\cfrac{3^(10)}{2^(16)}\cdot \cfrac{1}{2^(12)\cdot 3^8}\cdot 2^(28)\implies \cfrac{3^(10)\cdot 2^(28)}{2^(16)\cdot 2^(12)\cdot 3^8} \\\\\\ \cfrac{3^(10)\cdot 2^(28)}{2^(16+12)\cdot 3^8}\implies \cfrac{3^(10)\cdot 2^(28)}{2^(28)\cdot 3^8}\implies 3^(10)\cdot 2^(28)\cdot 2^(-28)\cdot 3^(-8)`


`\bf 3^(10-8)\cdot 2^(28-28)\implies 3^2\cdot 1\implies 3^2\implies 9`