Question

Question in comments, please help.

Answer 1

If a^m * a^n = a^mn then find the value of m(n-2) + n(m-2).

Answer 2

a^m * a^n should equal a^(m+n), not a^mn

Answer 3

I know, if it was a^(m+n), then it is one of the laws of exponents and all values of m and n will satisfy the condition. They are asking two numbers m and n where m+n=mn

Answer 4

\[ a^{(m+n - mn)} = a^0 \]
Squaring both sides,
\[ a^{2(m+n) - 2mn} = a^{2*0} = a^0 \]

Rest is easy!!

Answer 5

So,
2(m+n) - 2mn = 0
Am I right?

Answer 6

\[ m(n - 2) + n(m-2) = mn - 2m + mn - 2n = 2mn - 2(m+n) \]

Answer 7

I think so..

Answer 8

But there are two variables ... don't you think trial ad error would be easier?

Answer 9

*trial and error.

Answer 10

Please help @experimentX

Answer 11

you know that m + n = mn right? so do the distributive property on the original question and substitute back in with m + n = mn.

Answer 12

if we do the distributive property we end up with mn - 2m + nm -2n, now just put in m + n wherever you see mn. and you get zero.

Answer 13

You mean:
m+n-2m+m+n-2n
= 2m+2n-2m-2n
= 0
Thanks a lot!!