Question

Evaluat 1/2^-2x^-3y^5 for x=2 and y=-4

Answer 1

`-(1)/(32)`

Explanation:

Given Equation:

`(1)/(2^-^2*x^-^3*y^5)`

Multiplying numerator and denominator by
`2^2*x^3` gives:

`(2^2*x^3)/(y^5)`

We have to evaluate the equation at
`x=2 and y=-4`

Putting the values of x and y in the equation:

=
`(2^2*2^3)/((-4)^5)`

=
`(4*8)/((-1024))`

=
`-(32)/(1024) \\`

=
`-(1)/(32)`

On putting the value of 'x' and 'y' in the equation the solution of the equation is
`-(1)/(32)`

Answer 2

-1/32

Explanation:

1/(2⁻²x⁻³y⁵)

x=2

y=-4

1/(2⁻²x⁻³y⁵) = 1/(2⁻²2⁻³-4⁵)

4= 2²

equate it back where 4 is

1/(2⁻²2⁻³2²⁽⁵⁾) = 1/(2⁻²2⁻³2¹⁰)

using law of indices

1/(2⁻²⁺⁽⁻³⁾+¹⁰) = 1/(2⁻²⁻³⁺¹⁰) = 1/2⁻⁵⁺¹⁰= 1/2⁻⁵ = -1/2x2x2x2x2 = -1/32