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Question
Solve the following equation:
`8^(x+1)=16^(y+2)` and, `(1/2)^(3+x)=(1/4)^(3y)`
Solution
`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
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