Logarithmic Curves

\[y=\ln(ax+b),\qquad a\ne0\] Domain: \(ax+b>0\). The sketch illustrates the case \(a>0\), \(b>0\).

Vertical asymptote: \(ax+b=0\Rightarrow x=-\dfrac{b}{a}\).

\(x\)-intercept: \(y=0\Rightarrow ax+b=1\Rightarrow x=\dfrac{1-b}{a}\).

\(y\)-intercept: when \(b>0\), \(x=0\Rightarrow y=\ln b\). If \(b\le0\), there is no real \(y\)-intercept.