Question

Help me? idk the answer :P

Answer 1

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

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

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

$\bf 3^{10-8}\cdot 2^{28-28}\implies 3^2\cdot 1\implies 3^2\implies 9$