# College Algebra

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How would I go about solving for x in the equation:
ln(4x-2)-ln(4)=-ln(x-2)

• College Algebra -

ln[(4x-2)/4] = ln[1/(x-2)]
(4x -2)/4 = 1/(x-2)
x - 1/2 = 1/(x-2)
x^2 -x/2 -2x +1 = 1
x^2 -(5/2)x = 0
x = 0 or 5/2
0 is not allowed since ln of a negative number is undefined
x = 5/2

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