Question 1113137
The property of exponents  {{{ x^a/x^b  = x^(a-b) }}} (*)  will be used.   This can be easily proved.

Let  {{{log(b,(x/y)) = w }}}  <—>  {{{ b^w = x/y }}}  (1)
Let  {{{ log(b,(x)) = v }}}  <—>  {{{ b^v = x }}}   (2)
Let  {{{ log(b,(y)) = z }}}  <—>  {{{ b^z = y }}}  (3)

{{{ b^w = x/y = b^v/b^z = b^(v-z) }}}   (property (*) used on last step)

Picking off just this part:
{{{ x/y = b^(v-z) }}}

and taking log base b of both sides:
{{{ log(b,(x/y)) = log(b,(b^(v-z))) = v-z }}} 

Substituting (2) and (3) for v and z, respectively, completes the proof:
{{{ log(b,(x/y))= log(b,(x)) - log(b,(y)) }}}