# Solutions for logarithmic equation lnx+ln(x+1)=ln2

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We know that lnx +lny = lnxy

==> lnx+ln(x+1) =ln2

==> lnx(x+1) = ln2

==> x(x+1)=2

==> x^2+x-2=0

==> (x+2)(x-1)=0

==> x=1

( We will not consider x=-2 as a solution, because the function is not defined for negative values.)

To find the solutions to lnx+ln(x+1) = ln2.

Solution:

Since lna+lnb = ln(ab), we can write the equation as:

ln[x(x+1)] = ln2. Taking antilogarithm,

x(x+1) = 2.Or

x^2+x-2 = 0.

x^2+2x-(x+2) = 0

x(x+2)-1(x+2) = 0.

(x+2)(x-1) = 0. Or

x+2 = 0 . Or x-1 = 0.

x+2 = 0 gives: x = -2 not practical as ln is not defined for < 0 Or negative quantities.

x-1 = 0 gives: x= 1 which is the practical solution.