Question

Solve the following systems of equations.
$xy=a^2,(logx{)}^2+(logy{)}^2=\frac{5}{2}(log{a}^2{)}^2.$

Answer 1

Solution:Given,

=> xy = a²

=> x = a²/y equation → (1)

Now, (logx)² + (logy)² = 5/2( loga²)²

=> $\left(loga²/y\right)²+\left(logy\right)²=5/2\left(loga²\right)²$ [ • value of x from equation (1) ]

=> $(loga²-logy)²+\left(logy\right)²=5/2\left(loga²\right)²$

=>$\left(loga²\right)²+\left(logy\right)²-2loga²logy+\left(logy\right)²=5/2\left(loga²\right)²$

=>$\left(loga²\right)²+2\left(logy\right)²-2loga²logy-5/2\left(loga²\right)²=0$

=> $4\left(logy\right)²+2\left(loga²\right)²-5\left(loga²\right)²-4loga²logy=0$

Adding and subtracting ( loga²)² in RHS in above equation , we have

=> $\left(2logy\right)²+\left(loga²\right)²-4loga²logy-4\left(loga²\right)²=0$

=> $(2logy-loga²)²-4loga²=0$

=>$(2logy-loga²)²=\left(2loga²\right)²$

=>$(2logy-loga²)=\left(2loga²\right)$

=> $2logy=3loga²$

=> $logy²=log{a}^6$

=> $y²=a^6$ [∴ loga = logb ,a=b ]

=> $y=a³$ equation → (2)

And from equation (1) and (2) ,we have

=> $x=a²/y$

=> $x=a²/a³$

=> $x=a⁻¹$

=> $x=1/a$

• Therefore solution of systems of equations are X = 1/a ,Y = a³