Question 1176006
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Hi

log(x+2)+log(x-2)-log5 =0 

Log((x+2)(x-2)/5)= 0
(x+2)(x-2)/5) = 10^0 = 1
(x+2)(x-2) = 5
 x^2 - 4 = 5
  x^2 = 9
   x = ±3  Tossing out negative solution for a Logarithm
   x = 3

Log 1 = 0  checks
Wish You the Best in your Studies.
*[tex \large\ \ log_b(x) \ = \ y \ \ \Rightarrow\ \ b^y = x] ****
*[tex \large\ \ nlog_bx = log_b(x^n) ]
*[tex \large\ \ log_bx + log_by = log_b(xy) ] ****
*[tex \large\ \ log_bx - log_by = log_b(x/y) ] ****
*[tex \large\ \ log_b1 = 0]
*[tex \large\ \ log_bb = 1]
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