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

Guest Mar 27, 2018
 #1
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\(2\log{(a -2b)} = \log{a} + \log{b} \)

 

We can write

 

log (a - 2b)^2  = log ( a * b)

 

And we can solve

 

(a - 2b)^2  =  ab

a^2 - 4ab + 4b^2  = ab

a^2 - 5ab + 4b^2  = 0     factor

 

(a - 4b) ( a - b)  =  0

 

Setting  each factor to 0  and solving we have that

 

a  = 4b         and  a  = b

 

Assuming  a, b   are both positive, the first solution is the only one where all logs are defined for real numbers

 

So

 

a  / b  =

 

[4b]/ b  =

 

4

 

 

cool cool cool

CPhill  Mar 27, 2018

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