Just another way of writing it..

Hello,

as given ln x + ln (x+1) = 0,

ln x = -ln (x+1)

by exponenting both side,

e^( ln x ) = e^[ -ln (x+1) ]

e^(ln x) = e^[ ln(x + 1)^-1]

x = (x + 1)^-1

x = 1 / (x+1)

x(x + 1) = 1

x^(2) + x -1 = 0,

thanks yo....

$\begin{array}{l} ln(x) + \ln (x + 1) = 0\\ \Leftrightarrow \ln x(x + 1) = 0\\ \Leftrightarrow x(x + 1) = {e^0}\\ \Leftrightarrow x(x + 1) = 1\\ \Leftrightarrow {x^2} + x - 1 \end{array}$

x2+x+1 is solution solve it guyss

ln x + ln (x+1) = 0 It's actually ln a + ln b which is equal to ln ab Therefore, we get ln x + ln (x+1) = ln (x^2 + x) = 0 So, we can also write log to the base e ( x^2 + x) = 0 which can be written as x^2 + x = e^ 0 x^2 + x = 1 (anything to the power of 0 is 1) therefore, x^2 + x - 1 = 0