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[@bhargavclass7a129]

2^0 + 2^m = 2^(2m+3) + 2^(m+3) Case 1 On comparison if 2^0 = 2^m+3 Then 2^m=2^2m+3 So m+3=0, m=-3 2m+3=m, m=-3 Therefore one solution is m=-3 Case 2 2^2m+3=2^0, m=-3/2 Put -3/2 in 2^m+3 = 2^m then it doesn't satisfy this equation Therefore, the only solution is m=-3 m=-3

[@JeffreyLaryea]

I 😊love the question, and I was able to solve with log as well and I had -3🎉

[@meswag1233]

2^m = -1 has complex solutions e^(i*pi * (2c + 1)) / ln(2) for any integer c Great work though 👍

[@ElvisSaturn]

2^(2m+3) + 2^(m+3) = 2^m + 2⁰ with x=m+3 -> 2^m+x + 2^x = 2^m + 2⁰ -> x=0 -> m=-3

[@Kambyday]

(2^(m+3))(2^m + 1) = 2^m +2⁰ 2^(m+3) = (2^m + 1)/(2^m + 1) 2^(m+3) = 1 We know 2^(2niπ/ln(2)) = 1; n€Z So m+3 = 2niπ/ln(2) m = 2niπ/ln(2) - 3; n€Z Proof that 2^(2niπ/ln(2)) = 1 2^z = 1 z×ln(2) = ln(1) z = ln(e^2niπ)/ln(2) z = 2niπ/ln(2)

[@WorldwideBibleClass-qr9jk]

m=-3 I solved it in my head within 3eeconds. Why waste so much time solving this?

[@musicsubicandcebu1774]

You would only have to show your method in an exam.