Question 1045306
The expression {{{1/p+1/q-1/r-1/s}}} will be undefined if one or more of those denominators is zero.
If nothing was specified about the exponents,
we have to consider the case that one (or more) of them is zero:
If {{{pqrs=0}}} , then {{{1/p+1/q-1/r-1/s}}} is undefined.


If {{{pqrs<>0}}} , then {{{1/p+1/q-1/r-1/s=highlight(0)}}} :
Looking at just {{{c^r=d^s}}} , we can write
{{{c^r=d^s}}} ---> {{{(c^r)^(1/r)=(d^s)^(1/r)}}} ---> {{{c^"r/r"=d^"s/r"}}} ---> {{{c=d^"s/r"}}} .
Since {{{a^p=d^s}}} and {{{b^q=d^s}}} ,
we can also write {{{a}}} and {{{b}}} as powers of {{{d}}} :
{{{a=d^"s/p"}}} and {{{b=d^"s/q"}}} .
Substituting the expressions found for {{{a}}} , {{{b}}} , and {{{c}}} into {{{ab=cd}}} , we get to the solution.
{{{a*b=c*d}}} --> {{{d^"s/p"*d^"s/q"=d^"s/r"*d}}}  --> {{{d^"s/p"*d^"s/q"=d^"s/r"*d}}} --> {{{d^("s/p"+"s/q")=d^("s/r"+1)}}} --> {{{s/p+s/q=r/s+1}}}
And multiplkying both sides of the last equation above times {{{1/s}}} , we get
{{{1/p+1/q=1/r+1/s}}} .
So, subtracting {{{1/r+1/s}}} from both sides, we get
{{{1/p+1/q-(1/r+1/s)=1/r+1/s-(1/r+1/s)}}} --> {{{1/p+1/q-1/r-1/s=1/r+1/s-1/r-1/s}}} --> {{{highlight(1/p+1/q-1/r-1/s=0)}}} .