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the equation is

$\ln(x+2)=\ln e^{\ln2} - \ln x$

How do I solve for $x$?

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1  
$e^{\ln 2} = 2$, for starters –  The Chaz 2.0 Oct 23 '13 at 4:03
    
can you explain why ? –  user293849 Oct 23 '13 at 4:26
    
@user293849: Think about it: What does "$\ln 2$" mean? –  Blue Oct 23 '13 at 4:27
    
@user293849 Look up the logarithm identities $\ln a^b = b \ln a$. In your case $b=\ln 2$ and $\ln e=1$. Please have a look here en.wikipedia.org/wiki/List_of_logarithmic_identities –  triomphe Oct 23 '13 at 4:32

1 Answer 1

Your equation is equivalent to $$\ln (x+2)+\ln(x)=\ln2.$$ Then use logarithm identity $\ln ab=\ln a +\ln b$\$ $$\ln (x+2)x=\ln2.$$ Now take inverse of both sides $$(x+2)x=2.$$ Now you can solve the quadratic equation and select the appropriate $x$ value. You get $x=-1\pm\sqrt 3.$ Note that you can not take $x<0$ as $\ln x$ is not defined for negative $x$. So your answer is $x=-1+\sqrt 3$ which is approximately equal to $0.732$.

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