Q.

The equation loge x+log  (1+x)=0 can be written as

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a

x2+x-e=0

b

x2+x-1=0

c

x2+x+1=0

d

x2+xe-e=0

answer is B.

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Detailed Solution

logex+loge(1+x)=0⇒loge(1+x)=loge(1x) ⇒x(x+1)=1⇒x2+x-1=0
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The equation loge x+log  (1+x)=0 can be written as