Using the law of logarithm. LogA - LogB = Log(A/B)
log(x²-y²) - log(x-y) = log((x²-y²)/(x-y))
Note by difference of two squares, (x²-y²) = (x-y)(x+y)
Simplifying (x²-y²)/(x-y) = (x-y)(x+y)/(x-y) = (x+y)
Therefore log(x²-y²) - log(x-y) = log((x²-y²)/(x-y)) = log((x-y)(x+y)/(x-y)) = log(x+y)
