Solve the equation #2lnx-ln(x+1)=0#?

Question

2974 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-03T14:08:09

#x=(1+sqrt5)/2#

As #2lnx-ln(x+1)=0#, #x>0#

and we have #ln(x^2/(x+1))=0#

Hence #x^2/(x+1)=1#

or #x^2=x+1#

or #x^2-x-1=0#

or #x=(1+-sqrt(1^2+4))/2#

= #(1+-sqrt5)/2#

but as #x>0#, the only solution is #(1+sqrt5)/2~=1.618#

Check - #2ln1.618-ln2.618#

= #2xx0.48119-0.96241=0.96238-0.96241~=0#