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

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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**

ab+3a - 3b + 1 = 0

We'll factorize by 3:

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

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

Changing the base 72 of the number log 72 48 into the base 6, we'll get:

log 6 48 = log 72 48*log 6 72

But log 6 72 = log 6 (6*12)

We'll use 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.