Question

Solve x and y if : ( √32 )^x ÷ 2^{y + 1} = 1 and 8^y - 16^{4 - x/2} = 0

Answer 1

Consider the quation
`(sqrt32)^x ÷ 2^(y + 1) = 1`
⇒ `( sqrt(2 xx 2 xx 2 xx 2 xx 2))^x ÷ 2^(y + 1) = 1`
⇒ `( sqrt(2^5))^x ÷ 2^(y + 1) = 1`
⇒ `[(2^5)^(1/2)]^x ÷ 2^(y + 1) = x^0`

⇒ `2^(5x/2) ÷ 2^(y + 1 ) = x^0`

⇒ `(5x)/2 - ( y + 1) = 0`
⇒ 5x - 2( y + 1 ) = 0
⇒ 5x - 2y - 2 = 0                      ....(1)

Now consider the other equation
`8^y - 16^( 4 - x/2 ) = 0`
⇒ `(2^3)^y - (2^4)^(4 - x/2) = 0`

⇒ `2^(3y) - 2^[4( 4 - x/2)] = 0`

⇒ `2^(3y) = 2^[4( 4 - x/2)]`

⇒ `3y = 4( 4 - x/2)`

⇒ 3y = 16 - 2x
⇒ 2x + 3y = 16                      ...(2)

Thus, We have two equations,
5x - 2y = 2                           ...(1)
2x + 3y = 16                        ....(2)
Multiplying equation (1) by 3 and (2) by 2, We have
15x - 6y = 6                         ....(3)
4x + 6y = 32                        ....(4)
Adding equation (3) and (4), We have
19x = 38
⇒ x = 2 
Substituting the value of x in equation (1), We have
5(2) - 2y = 2
⇒ 10 - 2y = 2
⇒ 2y = 10 - 2
⇒ 2y = 8
⇒ y = `8/2`
⇒ y = 4
Thus the values of x and y are : x = 2 and y = 4.