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# Proving the equation: a(b+3) - 3b + 1 = 0 using logarithmic properties

Summary:

To prove the equation \(a(b+3) - 3b + 1 = 0\) using logarithmic properties, you must first isolate the logarithmic terms. Apply the properties of logarithms, such as the product, quotient, and power rules, to simplify the expression. Then, manipulate the equation algebraically to show that both sides are equal, confirming the equation's validity.

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

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

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math problem.log72 (6) + log 72 (8) = a and log6 4=b-1 . Prove that a(b+3)-3b+1=0.

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

ab + 3a - 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, ab + 3a - 3b + 1 = 0