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We know that lnx +lny = lnxy
==> lnx+ln(x+1) =ln2
==> lnx(x+1) = ln2
( 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.
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.
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