Hello,

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)