Winter Holiday

We have the expression $81^{-1/2} \times 81^{3/2}$ . Using the rules of exponents , we have the above expression as \[\begin{array}{} & = 81^{-1/2 +3/2} \\ & = 81^1 \\ & = 81. \square \end{array}\]

When you multiply two numbers with powers, you add their exponents. So this comes out to be 2/2 which is 81^1

just -1 over 2 + 3 over 2 which is 1 so the answer is just 81^1 which is 81

$81^{-\frac{1}{2}}\times81^{\frac{3}{2}} = 81^{-\frac{1}{2} + \frac{3}{2}}= 81^{\frac{2}{2}} = 81^1 = 81$

just -1 over 2 + 3 over 2 which is 1 so the answer is just 81^1 which is 81

3/2 - 1/2 = 2/2 = 1; 81^(1) = 81

81^-1/2 x 81^3/2

1/81^1/2 x (81^1/2)^3

1/9 x 9^3

9^2

81

81^-1/2 x 81^3/2

Take 81 common

81^3/2-1/2

81^2/2

81^1 = 81

There is a rule of multiplication , that is if same bases are multiplying , then their powers will be added, so if we add -1/2+3/2 it will be equals to 1 and any no. Whose power is 1 will remain same .

$81^{-1/2}\times81^{3/2} = 81^{3/2 - 1/2} = 81$

=(81^-1/2)*(81^3/2) =81^((-1/2)+(3/2)) =81

1/2 + 3/2 = 2/2 = 1 so 81^1 = 81

First(81^-1/2 )(81^3/2)=81^(-1/2+3/2)=81^2/2= 81^1=81

81^(3/2-1/2) = 81