Solve the following logarithmic equation in base 10 for \(x\) over the real numbers:
log \((2x-1)=\log\ (x+2)+\log\ (x-2)\)  3 (DO NOT ACCEPT 3 and -1) (Solution: \(2x-1=(x+2)(x-2)\); \((x-3)(x+1)=0\); \(x=3\), \(x eq\) -1 - the log(-1) does not exist in the real number system)