Question

Solve the following systems of equations.
$lo{g}_{2}x-lo{g}_{2}y+lo{g}_2(y+1)=1+lo{g}_{2}3,2^y{}^2-2^x.{16}^y=0.$

Answer 1

Solution: log₂(x) - log₂(y) + log(y + 1) = 1 + log₂(3) equation→(1)

=> $log₂\left(x\right)(y+1)/y=log₂\left(2\right)+log₂\left(3\right)$ [ ∴ loga + logb = logab ]

=> $log₂\left(x\right)(y+1)/y=log₂\left(6\right)$

=> $x(y+1)/y=6$ [∴ loga = logb , a = b ]

=> $x=6y/(y+1)$ equation → (2)

And, $\left(2\right)²ʸ-\left(2\right)ˣ\bullet \left(16\right)ʸ=0$ equation→(3)

=>$\left(2\right)²ʸ=\left(2\right)ˣ\bullet \left(2\right)⁴ʸ$

Taking log both side to the base 2,we have

=> log₂(2)²ʸ = log₂(2)ˣ ⁺ ⁴ʸ

=> 2ylog₂(2) = (x + 4y)log₂(2)

=> $2y=x+4y$

=> $x=-2y$

=> $6y/(y+1)=-2y$ [∴ value of X from equation→(2)]

=> $6y=-2y²-2y$

=> y(2y+2+6) = 0

=> $y=0$ equation→(4)

=> (2y + 8) = 0

=> $y=-4$ equation→(5)

Then,from equation (2),we have

=> $x=6(-4)/(-4+1)$ [∴ value of Y = - 4 from equation →(5)]

=> $x=8$

=> $x=6\left(0\right)/(0+1)$ [∴ value of Y = 0 from equation →(4) ]

=> $x=0$

But argument be always greater than and not equal to zero

so, y = -4,0 and x = 0 can not be solution of equation (1)

But for equation (3) solutions are :-

$X=0,8andY=0,-4$