What are the possible solutions of 2log2a = 10 – 2log2b, where a and b are integers? Show you’re calculations

Answer 1

2log2a = 10 – 2log2b
I will assume you mean:
2log₂a = 10 – 2log₂b
log₂ a^2 + log₂ b^2 = 10
log₂ (a^2 b^2) = 10
a^2 b^2 = 2^10
ab = 2^5 = 32
So integer pairs of factors of 32 will work
(a,b) = {(1,32), (2,16), (4,8), (8,4), (16,2), (32,1)}