How do you solve #ln (x + 2) + ln (x - 2) = 0#? Precalculus Properties of Logarithmic Functions Natural Logs 1 Answer Shwetank Mauria Jun 23, 2016 #x=sqrt5# Explanation: #ln(x+2)+ln(x-2)=0# is equivalent to #ln[(x+2)(x-2)]=ln1# or #(x+2)(x-2)=1# or #x^2-4=1# or #x^2-5=0# or #(x+sqrt5)(x-sqrt5)=0# or #x=-sqrt5# or #x=sqrt5# But as #x=-sqrt5# is not in domain (as log of negative number is not possible) hence #x=sqrt5# Answer link Related questions What is the natural log of e? What is the natural log of 2? How do I do natural logs on a TI-83? How do I find the natural log of a fraction? What is the natural log of 1? What is the natural log of infinity? Can I find the natural log of a negative number? How do I find a natural log without a calculator? How do I find the natural log of a given number by using a calculator? How do I do natural logs on a TI-84? See all questions in Natural Logs Impact of this question 1338 views around the world You can reuse this answer Creative Commons License