It is given that log(72) 48 = a and log(6) 24 = b

a = log(72) 48 = log(6) 48/ log(6) 72

=> log(6) (6*8)/log(6) 6*12

=> [1 + log(6) 8]/[1 + log(6) 12]

=> [1 + 3*log(6) 2]/[2 + log(6) 2]

b = log(6) 24 = log(6) 4*6 = 1 + log(6) 4 = 1 + 2*log(6) 2

Let log(6) 2 = x

=> a = [1 + 3x]/[2 + x]

=> b = 1 + 2x

a*(b + 3) - 3b + 1

(1 + 3x)(3 + 1 + 2x)/(2 + x) - 3(1 + 2x) + 1

=> (1 + 3x)(4 + 2x)/(2 + x) - 3(1 + 2x) + 1

=> (1 + 3x)*2 - 3(1 + 2x) + 1

=> 2 + 6x - 3 - 6x + 1

=> 0

This proves that if log(72) 48 = a and log(6) 24 = b, a*(b+3) - 3b + 1 = 0