Answer:
1). 
2). log(x - 1) + log5x = 2
3). ln(x + 5) = ln(x - 1) + ln(x + 1)
4).  
 
Step-by-step explanation:
1). ln(x + 5) = ln(x - 1) + ln(x + 1)
    ln(x + 5) = ln(x - 1)(x + 1)  [Since ln(a×b) = ln a + lnb]
    (x + 5) = (x- 1)(x + 1)
    x + 5 = x² - 1
    x² - x - 6 = 0
    x² - 3x + 2x - 6 = 0
    x(x - 3) + 2(x - 3) = 0
    (x + 2)(x - 3 ) = 0
    x = -2, 3
    But x = -2 is an extraneous solution.
    Therefore, x = 3 is the only solution.
2). 
   x² = 4x +5
   x² - 4x - 5 = 0
   x² - 5x + x - 5 = 0
   x(x - 5) + 1(x - 5) = 0
   (x + 1)(x - 5) = 0
   x = -1, 5
   Therefore, solution set is (-1, 5)
3). 
    5x² + 2 = (x + 8)  
    5x² - x - 6 = 0
    5x² - 6x + 5x - 6 = 0
    x(5x - 6) + 1(5x - 6) = 0
    (x + 1)(5x - 6) = 0
    x = -1, 
4). log(x - 1) + log5x = 2
     log(x - 1)(5x) = 2
     5x(x - 1) = 10² [if loga = b,  ]
]
     5x² - 5x - 100 = 0
     x² - x - 20 = 0
     x² - 5x + 4x - 20 = 0
     x(x - 5) + 4(x - 5) =0
     (x - 5)(x + 4) = 0
     x = -4, 5
But x = -4 is an extraneous solution.
Therefore, x = 5 is the only solution.