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
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
I 😊love the question, and I was able to solve with log as well and I had -3🎉
2^m = -1 has complex solutions
e^(i*pi * (2c + 1)) / ln(2) for any integer c
Great work though 👍
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
(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)
m=-3 I solved it in my head within 3eeconds. Why waste so much time solving this?
You would only have to show your method in an exam.