Question

$${ \infty }^{ 2+3 } \times { \infty }^{ 9+8 } =$$

Answer

$$o^44$$

Solution


Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[{({o}^{2})}^{2+3}{({o}^{2})}^{9+8}\]
Simplify  \(2+3\)  to  \(5\).
\[{({o}^{2})}^{5}{({o}^{2})}^{9+8}\]
Simplify  \(9+8\)  to  \(17\).
\[{({o}^{2})}^{5}{({o}^{2})}^{17}\]
Use Power Rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[{o}^{10}{({o}^{2})}^{17}\]
Use Power Rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[{o}^{10}{o}^{34}\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[{o}^{10+34}\]
Simplify  \(10+34\)  to  \(44\).
\[{o}^{44}\]