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20/2^5=1/2^m+1/2^n,where m and n are integers, mn=?

mn=3, is it right? Assume that m>n 5 x 2^(m+n)= 8 x 2^n x[2^(m-n)+1] So 2^(m-n)+1=5 So m-n=2 => m=n+2 Then 5*26(2n+2)=8*2^n*5 => 2n+2=n+3 Then n=1, m=2

20/2^5=1/2^m+1/2^n,where m and n are integers, mn=?

mn=3, is it right? Assume that m>n 5 x 2^(m+n)= 8 x 2^n x[2^(m-n)+1] So 2^(m-n)+1=5 So m-n=2 => m=n+2 Then 5*26(2n+2)=8*2^n*5 => 2n+2=n+3 Then n=1, m=2

Here is my question. I simplified it down to 2^m + 2^n / 2^m+n = 5/8 ... then i equated the numerators to each other, and the denominators to each other, i.e.:

2^m + 2^n = 5 2^m+n = 8

this gave me two different answers .... where did i go wrong ?

My three goals of business school: entrepreneurship, network, and professor mentor. I want to build something. I want to meet new people and create life-long friendships. I want to...