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How do you solve #ln x + ln (x-2) = 1#?

1 Answer

May 20, 2016

#x = 1 + sqrt[1 + e]#

Explanation:

From #log_e x + log_e(x-2) = log_e(e)# we conclude
#log_e(x(x-2))=log_e(e)# and also
#x(x-2)=e#. Solving for #x# we obtain
#x = 1 + sqrt[1 + e]#

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