Question 437728
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Hi
*[tex \LARGE\ \ log_2(7+2x) + log_2(4x-1) = log_2(2x+7)(4x-1) = 3 ]
*[tex \LARGE \ \ \ \ \ \ log_b(x) \ = \ y \ \ \Rightarrow\ \ b^y = x]
  2^3 = (2x+7)(4x-1)
    8x^2 +26x -15 = 0
factoring
 (4x +15)(2x -1) = 0
 (4x +15)= 0
     x = -15/4 Extraneous solution: (7+2x) and (4x-1) are < 0 
  (2x -1) = 0
    x = 1/2