#1**+1 **

2log(a-2b) = loga + logb so log(a-2b)^{2} = logab

therefore (a-2b)^{2} = ab

expanding : a^{2} -4ab + 4b^{2} = ab

ie a^{2} - 5ab + 4b^{2} = 0

divide both sides by b^{2} : a^{2}/b^{2} -5ab/b^{2} + 4 =0

ie (a/b)^{2 }- 5(a/b) + 4 = 0

solving the quadratic for a/b gives (a/b - 1)(a/b - 4) = 0

therefore a/b =1 or a/b = 4

Guest Feb 23, 2018

#2**+1 **

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

log (a - 2b)^2 = log (a*b)

This implies that

(a - 2b)^2 = ab

a^2 - 4ab + 4b^2 = ab

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

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

This implies that

a = 4b ⇒ a/b = 4 or

a - b = 0 ⇒ a = b ⇒ a/b = 1

CPhill
Feb 23, 2018

#3**+1 **

Wondering if a/b =1 might be considered an extraneous (invalid) solution, as it would result in a

LOG (negative number) ??

ElectricPavlov
Feb 23, 2018

#4**0 **

No I don't think so as squaring a negative gives a positive

ie 2log(a-2b) =log (a-2b)^{2}

Guest Feb 23, 2018

#5**+1 **

That may be true, but on the left you still have a LOG (negative number) .....So I am confused.

2 Log (a-2b) = log(a-2b)^2

seems invalid as the left side of the equation is invalid if (a-2b) is negative

Does anyone know for SURE?

Help !

~EP

ElectricPavlov
Feb 23, 2018

#6**0 **

Perhaps the author of the question didn't believe it an issue otherwise they would have included that a-2b >= 0 in the question. I do see your dilemma but if you remember back to when you learnt all about logs and I guess it depends on how your teacher treated the topic. A logarithm as we know is the power of a base to obtain a given number. From this definition we get a whole new mathematics topic part of which is simplifications. The actual log law in question is that if we have a log of a power then : log_{b}x^{n} = nlog_{b}x not the other way round. Of course the other way has been adopted also to help with simplification. Maybe there are those who think its a case of which came first the chicken or the egg?

Guest Feb 23, 2018

#7**+1 **

The chicken ! No wait...... The egg! No, wait.....

Thanx for the clarification help !

~EP

ElectricPavlov
Feb 23, 2018