The equation loge x+ loge (1+x)=0 can be...

The equation `log_e x+ log_e (1+x)=0` can be written as

A

` x^2 +x-e=0`

B

`x^2 +x-1=0`

C

`x^2+x +1=0`

D

`x^2 + xe -e=0`

Verified by Experts

The correct Answer is:

B

Similar Questions

Explore conceptually related problems

If log e^(x)+log e^(1+x)=0 then x=

Solve : log_e (x-a) + log_e x =1 (a>0)

The number of real solution(s) of the equation 9^(log_(3)(log_(e )x))=log_(e )x-(log_(e )x)^(2)+1 is equal to

The particular solution of the differential equation y(1+log x)(dx)/(dy)-x log x=0 when x=e,y=e^(2) is

The domain of f(x)=log_(e)|log_(e)^(x)| is

For the function f(x)=(log_(e)(1+x)+log_(e)(1-x))/(x) to be continuous at x = 0, the value of f(0) is

For the function f(x) = (log_(e )(1+x)-log_(e )(1-x))/(x) to be continuous at x = 0, the value of f(0) should be

If a,b and c are positive numbers in a GP,then the roots of the quadratic equation (log_(e)a)^(2)-(2log_(e)b)x+(log_(e)c)=0 are