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Find a/b when 2log{(a -2b)} = log{a} + log{b}.

Guest May 30, 2018
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Solve for b:

2 log(a - 2 b) = log(a) + log(b)

 

Subtract log(a) + log(b) from both sides:

-log(a) + 2 log(a - 2 b) - log(b) = 0

 

-log(a) + 2 log(a - 2 b) - log(b) = log(1/a) + log((a - 2 b)^2) + log(1/b) = log((a - 2 b)^2/(a b)):

log((a - 2 b)^2/(a b)) = 0

 

Cancel logarithms by taking exp of both sides:

(a - 2 b)^2/(a b) = 1

 

Multiply both sides by a b:

(a - 2 b)^2 = a b

 

Subtract a b from both sides:

(a - 2 b)^2 - a b = 0

 

Expand out terms of the left hand side:

a^2 - 5 a b + 4 b^2 = 0

 

The left hand side factors into a product with two terms:

(a - 4 b) (a - b) = 0

Split into two equations:

a - 4 b = 0 or a - b = 0

 

Subtract a from both sides:

-4 b = -a or a - b = 0

 

Divide both sides by -4:

b = a/4 or a - b = 0

 

Subtract a from both sides:

b = a/4 or -b = -a

 

Multiply both sides by -1:

 

b = a/4    So, a / b = 4

Guest May 30, 2018
edited by Guest  May 30, 2018

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