Question

Please help!! 100 points! Help!

Answer 1

hope this helps .....

Answer 2

log² ( 1 + 4/x ) + log²{ 1 - 4/(x + 4)} = 2log² { 2/(x -1) -1 }

log² ( 1 + 4/x ) + log²{( x +4 - 4 )/(x + 4) }= 2log² { 2 - ( x -1) }/(x -1) }

log² ( 1 + 4/x ) + log²{x/( x + 4) } = 2log² { (3-x)/( x - 1) }

log²( 1 + 4/x ) + log²( 1/( 1 + 4/x ) } = 2log² (3 - x )/(x -1)


log²(1 +4/x ) + { -log(1 + 4/x) }² =2 log² (3 -x)/(x -1)

log²( 1 + 4/x ) + log²( 1+ 4/x ) = 2log²(3 -x)/(x -1)

2log²( 1 + 4/x ) = 2log²( 3 -x )/(x -1)

{ log( 1+ 4/x ) }² = { log(3 -x )/(x -1)} ²

log( 1+ 4/x ) = ± log(3 -x )/( x -1)

log(1 +4/x ) = log(3 -x )/(x -1)

(1 + 4/x ) = (3 -x )/( x -1)

(x +4)(x -1)= 3x -x²

x² +4x -x -4 = 3x -x²

2x² - 4 = 0
x = ± √2

again,
(x + 4) ( x -1) = - x ( 3 - x )

x² +3x -4 = -3x + x²
6x = 4

x = 2/3

now, for log to be defined ,
(1 + 4/x ) > 0 ,=> x > 0 , x < -4
(3 - x)/(x -1) > 0 => 1 < x < 3
if you see wave curve line then,
common interval is
1 < x < 3

so, x = √2

hence, answer is x = √2