Question

Logx-logx-1^2=2log (x-1)

Answer 1

Answer:

MARK AS BRAINLIEST

THANK ME

FOR INBOX

Step-by-step explanation:

Logx - 1/2 log( x -1/2) = log( x + 1/2) - 1/2log( x + 1/8)

logx - log( x -1/2)½ = log( x + 1/2) -log( x + 1/8)½

logx - log( x + 1/2) = log( x -1/2)½ - log( x + 1/8)½

logx/( x + 1/2) = log{ ( x -1/2)/( x + 1/8)}½

x/( x + 1/2) = √(x -1/2)/√(x + 1/8)

take square both sides

x²/( x + 1/2)² = ( x - 1/2)/( x + 1/8)

x²(x + 1/8) = (x+ 1/2)²( x - 1/2)

x³ + 1/8x = (x² - 1/4)(x +1/2) = x³ +1/2x² -1/4x -1/8

1/8x = 1/2x² - 1/4x -1/8

1/8x + 1/4x = 1/2x² -1/8

3/8x = 4/8x² -1/8

4x² - 3x - 1 = 0

4x² -4x + x -1 = 0

4x( x -1) +1( x -1) = 0

x = -1/4 and 1

now for log to be defined

x >0

x > 1/2

x > -1/2

x > -1/8

so, common value of this

x > 1/2

so, x ≠ -1/2

hence, x = 1