Question

Log (3x-1) + log (3x+1) = log 16

Answer 1

Answer:

We can write log[(3x-1)(3x+1)] = Log (3x-1) + log (3x+1)

Step-by-step explanation:

hence log[(3x-1)(3x+1)] = log16

taking anti log both sides

9x^2 - 1 = 16

x^2 = (16+1)/9

x^2 = 17/9

x = +√17/3 and x = -√17/3