Question 969493
x=5

Start by rewriting as Log (b2) {(x+3)(x-3)}=4.  Adding logs is like multiplying them with the same base.  

Think of log 10 +log 100=1+2=3;  log 10*100=log 1000=3.

From the definitions of logs, 2^4=16=(x+3)(x-3)=x^2-9.

So, x^2-9=16; x^2=25; x+ 5 or -5.  Negative logs don't exist, so -5 is extraneous.

Try 5 in the original
Log b2 (8) +log b2 (2)=4 ?
The first is 3 (2^3=8)  and the second is 1; 3+1=4.