SOLUTION: If {{{a^m*a^n=a^mn}}},then m(n-2)+n(m-2)is?options are (a)-1,(b)1 (c)0 (d)1/2

Algebra ->  Expressions-with-variables -> SOLUTION: If {{{a^m*a^n=a^mn}}},then m(n-2)+n(m-2)is?options are (a)-1,(b)1 (c)0 (d)1/2      Log On


   

Question 458405: If a%5Em%2Aa%5En=a%5Emn,then m(n-2)+n(m-2)is?options are (a)-1,(b)1 (c)0 (d)1/2
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Use the idea that x%5E%28y%29%2Ax%5E%28z%29=x%5E%28y%2Bz%29 to state that a%5Em%2Aa%5En=a%5Emn is also the same as a%5E%28m%2Bn%29=a%5Emn


Now, if the bases are equal, then the exponents are equal. So m%2Bn=mn or mn=m%2Bn


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%2Bn)


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%5Em%2Aa%5En=a%5Emn



So the answer is choice C)