Question 458405
Use the idea that {{{x^(y)*x^(z)=x^(y+z)}}} to state that {{{a^m*a^n=a^mn}}} is also the same as {{{a^(m+n)=a^mn}}}



Now, if the bases are equal, then the exponents are equal. So {{{m+n=mn}}} or {{{mn=m+n}}}



Now all we need to do is start with m(n-2)+n(m-2) and simplify



m(n-2)+n(m-2)



mn-2m+nm-2n 



mn-2m+mn-2n 



mn+mn-2m-2n 



2mn-2(m+n)



2(m+n)-2(m+n) .... Note: I've replaced mn with m+n (since {{{mn=m+n}}})



0



So all this shows us that m(n-2)+n(m-2) = 0 when m+n = mn




Consequently, this means that m(n-2)+n(m-2) = 0 if {{{a^m*a^n=a^mn}}}




So the answer is choice C)