Question 415755
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Hi
*[tex \LARGE\ \ 2log_3(x+4) - log_39 = log_3(x+4)^2/9 = 2 ]

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ log_b(x) \ = \ y \ \ \Rightarrow\ \ b^y = x]
  3^2 = (x+4)^2/9
  81= (x+4)^2
 ± 9 = x + 4
  -4 ± 9 = x   (x = -13 is an Extraneous solution)
   x = 5