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# Logarithm Problem

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find $$\frac{a}{b}$$ when$$2 log(a-2b) = log(a) + log(b)$$

Mar 17, 2019

#1
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2 log ( a - 2b)   =  log (a) + log(b)

log ( a - 2b)^2  = log (ab)      .....so....

(a - 2b)^2  = ab

a^2 - 4ab +4b^2  =  ab

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

(a - 4b) ( a - b)  = 0

Setting each factor to 0    and solving we have that

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

a = 4b                                 a = b

a/b = 4                                a/b = 1

The second solution is no good because  if  a = b......then     a - 2b  =  a - 2a   =  -a

Which means that log ( a - 2b) = log (-a)    and is defined only if a is negative

But....on the right side of the original equation , log (a)   produces a log of a negative which is not defined for real numbers

So

a / b  = 4   Mar 17, 2019
#2
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ok I got down to as much as the log(a-2b)^2 = log(ab), but the part I'm confused about is why if there are two logs on both sides you can cancel them out?

EDIT: No wait, I just realized since they're the same base lol, I was being dumb

Guest Mar 17, 2019
edited by Guest  Mar 17, 2019
#3
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HAHAHA!!!!!......you're not "dumb".....you just forgot.....   CPhill  Mar 17, 2019