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I have to solve : $$0 = \ln(x) - \ln(1-x).$$

I did:

$0 = \ln(x) - \ln(1-x) \implies e^0 = e^{\ln(x)} - e^{\ln(1-x)} \implies 1 = x - (1-x) \implies 2 = 2x \implies x=1 $

The solution should be $x=\frac{1}{2}$

Where is my fault ? :(

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2 Answers

up vote 9 down vote accepted

After you take '$e$' of both sides you do indeed arrive at $$ e^0 = e^{ \ln x - \ln(1-x) } $$ because of the general property that $ a=b \implies e^a=e^b$. However, you then tried to use a false rule, that $e^{a-b} = e^{a} - e^{b} $. If you recall your index laws, you'll remember that the proper rule is $$ e^{a-b} = \frac{e^a}{e^b} .$$

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The problem is that if you take exponentials of both sides, you should get $$e^0 = e^{\ln(x) - \ln(1-x)}.$$ But $$e^{\ln(x) - \ln(1-x)} \neq e^{\ln(x)} - e^{\ln(1-x)}.$$ So your first step is incorrect.

Simplest thing is to first go from $0 = \ln(x) - \ln(1-x)$ to $\ln(1-x) = \ln(x)$ and then take the exponentials.

Alternatively, from $$ e^0 = e^{\ln (x) - \ln(1-x)}$$ you can use the fact that $e^{a-b} = \frac{e^a}{e^b}$ to get $$ 1 = \frac{e^{\ln(x)}}{e^{\ln(1-x)}},$$ and proceed from there.

P.S. Please don't use [homework] as the only tag in your post.

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Maybe we could do a feature request that would require an additional tag besides "homework"... –  The Chaz 2.0 Oct 10 '11 at 16:51
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it's my third post here. so bear with me :( .. i will keep it in mind (regarding the homework-tag) –  fragant1996 Oct 10 '11 at 16:54
    
@fragant1996: It's not a big deal, but please do keep it in mind going forward. –  Arturo Magidin Oct 10 '11 at 17:06
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