If log72 48 = a and log6 24 = b prove that a(b+3) - 3b + 1 = 0

We'll remove the brackets from the expression in a and
b:

ab+3a - 3b + 1 = 0

We'll
factorize by 3:

ab + 3(a-b) + 1 = 0
(*)

We'll substitute a and b by the given
logarithms:

log 72 48*log 6 24 + 3(log 72 48 - log 6 24) +
1 = 0

We'll change the base 72 of the number log 72 48 into
the base 6:

log 6 48 = log 72 48*log 6
72

But log 6 72 = log 6
(6*12)

We'll apply the rule of
product:

log 6 (6*12) = log 6 6 + log 6
12

log 6 (6*12) = 1 + log 6
12

But log 6 12 = log 6
(6*2)

log 6 (6*2) = log 6 6 + log 6
2

log 6 (6*2) = 1 + log 6
2

log 6 48 = log 72 48*(2 + log 6 2)
(1)

We'll write log 6 48 = log 6 6 + log 6
8

log 6 48 = 1 + 3log 6 2
(2)

We'll substitute (2) in
(1):

1 + 3log 6 2 = log 72 48*(2 + log 6
2)

We'll divide by (2 + log 6
2):

log 72 48 = (1 + 3log 6 2)/(2 + log 6 2)
(3)

We'll also write log 6
24:

log 6 24 = log 6 6 + 2log 6
2

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

We'll substitute (3) and (4) in
(*):

(1 + 3log 6 2)(1 + 2log 6 2)/(2 + log 6 2) + 3[(1 +
3log 6 2)/(2 + log 6 2) - 1 - 2log 6 2] + 1 = 0

We'll
remove the brackets and we'll multiply by (2 + log 6 2):

1
+ 2log 6 2 + 3log 6 2 + 6(log 6 2)^2 - 3 - 6log 6 2 - 6(log 6 2)^2 + 2 + log 6 2 =
0

We'll combine and eliminate like terms and we'll
get:

0 = 0
true

So, for log 72 48 = a and
log 6 24 = b, the equality

ab+3a - 3b + 1 = 0 is
true.