Question

loge x + log(1+x) = 0 is equivalent to
(a) x2+x+e = 0
(b) x2+x-e = 0
(c) x2+x+1 = 0
(d) x2+x-1 = 0 ​

Answer 1

Step-by-step explanation:

Correct option is

B

x

2

+x−1=0

log

e

x+log

e

(1+x)=0

log

e

x(1+x)=0

x(x+1)=0

x

2

+x=1

x

2

+x−1=0