As the domain for lnxlnx is x>0x>0, the domain for ln(1+x)ln(1+x) is x> -1x>−1. However range is -oo < ln(1+x) < oo−∞<ln(1+x)<∞.
Function is continuous and relation is one to one.
Again for 0 < x < 10<x<1, lnxlnx is negative and hence ln(1+x)ln(1+x) is negative for -1 < x < 0−1<x<0.
And as x->-1x→−1, ln(1+x)->-ooln(1+x)→−∞ and hence x+1=0x+1=0 is asymptote for ln(1+x)ln(1+x).
At x=0x=0, ln(1+x)=0ln(1+x)=0 (it also means ln(1+x)ln(1+x) cuts xx-axis at (0.0)(0.0) and for x>0x>0 ln(1+x)>0ln(1+x)>0.
One can also take additional values for xx such as {1,2,3,4,5,6,7,8,9,10}{1,2,3,4,5,6,7,8,9,10} for which ln(1+x)ln(1+x) would be {0.693,1.099,1.386,1.609,1.792,1.946,2.079,2.197,2.303,2.398,2.485}{0.693,1.099,1.386,1.609,1.792,1.946,2.079,2.197,2.303,2.398,2.485}. The graph of ln(1+x)ln(1+x) appears as given below.
graph{ln(1+x) [-5.415, 14.585, -6, 4]}
