Question

7^16/7^12=7^-18/? what is the missing denominator

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}}}}\\\\ -------------------------------\\\\`


`\bf \cfrac{7^(16)}{7^(12)}=\cfrac{7^(-18)}{x}\implies x\cdot 7^(16)=7^(12)\cdot 7^(-18)\implies x\cdot 7^(16)=7^(12-18) \\\\\\ x\cdot 7^(16)=7^(-6)\implies x=\cfrac{7^(-6)}{7^(16)}\implies x=\cfrac{7^(-6)\cdot 7^(-16)}{1}\implies x=7^(-6-16) \\\\\\ \boxed{x=7^(-22)}\implies x=\cfrac{1}{7^(22)}`