Question

Solve the following equation:

`8^(x+1)=16^(y+2)` and, `(1/2)^(3+x)=(1/4)^(3y)`

Answer 1

`8^(x+1)=16^(y+2)` and, `(1/2)^(3+x)=(1/4)^(3y)`

`rArr(2^3)^(x+1)=(2^4)^(y+2)` and `(1/2)^(3+x)=(1/2^2)^(3y)`

`rArr(2)^(3x+3)=(2)^(4y+8)` and `(1/2)^(3+x)=(1/2)^(6y)`

⇒ 3x + 3 = 4y + 8 and 3 + x = 6y

⇒ 3x - 4y = 8 - 3

⇒ 3x - 4y = 5  ...........(i)

Now,

3 + x = 6y

x = 6y - 3             ..............(ii)

Putting x = 6y - 3 in equation (i), we get

3(6y - 3) - 4y = 5

⇒ 18y - 9 - 4y = 5

⇒ 14y = 14

⇒ y = 1

Putting y = 1 in equation (ii) we get,

x = 6(1) - 3 = 3