Question

Solve the following systems of equations.
$2lo{g}_4(y+1)+lo{g}_{2}y=lo{g}_2(\frac{x}{y}-2),5+lo{g}_2\frac{y}{x}=\frac{6}{lo{g}_2\frac{x}{y}}.$

Answer 1

Solution: Given,

=> 5 + log₂(y/x) = 6/log₂(x/y)

=> 5 + log₂(y/x) = -6/log₂(y/x) [ ∴ log₂(x/y) = log₂(y/x)⁻¹ and log₂(y/x) = -1log₂(y/x) ]

• let log₂(y/x) = t

Then, 5 + t = -6/t

=> 5t + t² = -6

=> t² + 5t + 6 = 0

=> t² + 3t + 2t + 6 = 0

=> t(t + 3) + 2(t + 3) = 0

=> (t + 3)(t + 2) = 0

=> t = -2 , => t = -3

• As, log₂(y/x) = t

=> log₂(y/x) = -2 and => log₂y/x = -3

=> y/x = 2⁻² and => y/x = 2⁻³ [∴ loga = logb , a = b]

=> x = 4y and => x = 8y equation→(1)

• And, 2log₄(y+1) + log₂y = log₂(x/y - 2)

=> 2/2log₂(y + 1) + log₂y = log₂(x/y - 2) [∴ logₐⁿ b = 1/nlogb ]

=> log₂(y + 1)y = log₂(x/y - 2) [∴ loga + logb = logab ]

=> (y + 1)y = x/y - 2 [∴ loga = logb , a = b]

=> y² + y = (x - 2y)/y

=> y³ + y² = x - 2y

• Putting the value of X from equation →(1) ,we have

=> y³ + y² + 2y = 4y and => y³ + y² + 2y = 8y

=> y³ + y² - 2y = 0 and => y³ + y² - 6y = 0

• These are cubic equations ,then solutions are as follow:-

=> y₁ = 1 and => y₄ = 2

=> y₂ = -2 and => y₅ = -3

=> y₃ = 0 and => y₆ = 0

• So, argument of log canot be zero and negative .

• Then, solutions are y = 1 , 2

• From equation (1),we have

=> x = 4y and => x = 8y

• for y = 1

=> x = 4 and => x = 8

• for y = 2

=> x = 8 and => x = 16

• Therefore,solutions of systems of equations are

X = 4,8,16 and Y = 1,2