0 votes 0 votes If $\log _{e}x+\log _{e}(1+x)=0,$ then: $x^{2}+x-1=0$ $x^{2}+x+1=0$ $x^{2}+x-e=0$ $x^{2}+x+e=0$ Quantitative Aptitude nielit2019feb-scientistc logarithms + – Lakshman Bhaiya asked Apr 1, 2020 • edited May 17, 2020 by go_editor Lakshman Bhaiya 13.8k points 654 views answer comment Share See 1 comment See all 1 1 comment reply haralk10 778 points commented Apr 21, 2021 reply Share option (A) 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes option A is right s_dr_13 answered Mar 27, 2022 s_dr_13 228 points comment Share See all 0 reply Please log in or register to add a comment.