Question

loge x +loge(1+x) =0. then x²+x=?​

Answer 1

Answer:

$x^{2} + x = 1$

Step-by-step explanation:

$loge(x) + loge(1+x) = 0$

$loge[x(1+x)] =0 \\ {loge(m)+loge(n) = loge(m•n)}$

$loge(x+x^{2} ) = 0$

$(x+x^2) = 1$

$because \: - > loge(1) = 0$