If there is no mistake in the subject, i'll write this :
So you have to solve :
Differentiate the case when ln(x²-3) is positive (ie x²-3 > 1) and the case when ln(x²-3) is negative (ie x²-3 < 1).
Multiply the two sides with ln(x²-3), change the sign when it's negative.
Then, you may have ln(4x+2) > (or <) ln(x²-3), which is ln(4x+2) - ln(x²-3) > (or <) 0=ln(1)
<=> (or <)
Ln is an increasing function, so a<b <=> ln(a)<ln(b)