Step 1: Apply the logarithm property lna+lnb=ln(ab) for a>0, b>0.
ln[(x+1)(x−1)]=ln3
Step 2: Since ln is one-to-one, equate the arguments.
(x+1)(x−1)=3
Step 3: Expand the left side.
x2−1=3
Step 4: Solve for x2.
x2=4
x=±2
Step 5: Determine the domain where both arguments are positive: x+1>0 and x−1>0.
x>1
Step 6: Check candidate solutions in the domain.
For x=2>1: valid.
Verify: ln(3)+ln(1)=ln3+0=ln3.
For x=−2<1: ln(−1) undefined in reals. Invalid.
Final answer
x=2