Question

( log _{e} x+log _{e}(1+x)=0 )
( log _{e} x(1+x)=0 )
( operatorname{lye} x(1+x)=operatorname{ly}_{e^{1}} )
( x^{2}+x-1=0 )

# 26. The equation loge * +loge (1+x)=0 can be written as (a) x²+x-1=0 (b) er? +x+1=0 (c) x²+x-e=0 (d) x +x+e=0.

Solution