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

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### 2 Answers

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

Solution:

We convert the logarithms to base 6.

log 72 (48) = log6(48)/log6(722) = log6 (6*8)/log6 (6^2*2).

log 72 (48) = {1+log6 (2^3}/{2+log2} = {1+3log6 (2)}/{(2+log6(2)}.

Therefore ,

log 72 (48) = a = {1+3log6 (2)}/{2+log6 (2)}.....(1).

Also log6 (24) = log6 {6*2^2} = {1+ log6 (2^2)} = {1+2log 6 (2)}....(2).

From eq (2), we get log6(2) = (b-1)/2. We substitute this in eq (1) and get:

a = {1+3 (b-1)/2}/{2+(b-1)/2).

a = {2+3(b-1)}/{4+b-1)} = (3b+1)/(b+3).

a(b+3) = (3b+1).

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.**