# How do I solve this logarithmic equation?

the equation is

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

How do I solve for $x$?

-
$e^{\ln 2} = 2$, 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

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\$.