Solve for x:
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concepts :
.....(1)
.........(2)
now, log(3x + 2) - log(3x - 2) = log5
or, log{ (3x + 2)/(3x - 2) } = log5 [ using formula (1) ]
or, (3x + 2)/(3x - 2) = 5 [ using formula (2), ]
or, (3x + 2) = 5(3x - 2)
or, 3x + 2 = 15x - 10
or, 3x - 15x = -10 - 2
or, -12x = -12
hence, x = 1
now, log(3x + 2) - log(3x - 2) = log5
or, log{ (3x + 2)/(3x - 2) } = log5 [ using formula (1) ]
or, (3x + 2)/(3x - 2) = 5 [ using formula (2), ]
or, (3x + 2) = 5(3x - 2)
or, 3x + 2 = 15x - 10
or, 3x - 15x = -10 - 2
or, -12x = -12
hence, x = 1
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