The natural logarithm ln xEdexcel International A Level Maths: Revision notes

The natural logarithm $\ln x$ is the logarithm to base $e$: $y=\ln x$ means $x=e^{y}$. Its domain is $x>0$ and its range is all real numbers. The graph of $y=\ln x$ passes through $(1,0)$, increases for all $x>0$ (slowly for large $x$), and has the $y$-axis ($x=0$) as a vertical asymptote: $\ln x\to-\infty$ as $x\to0^{+}$. Also $\ln e=1$ and $\ln1=0$.

The functions $e^{x}$ and $\ln x$ are inverses: $\ln(e^{x})=x$ for all real $x$ and $e^{\ln x}=x$ for $x>0$. Their graphs are reflections of each other in the line $y=x$, so the asymptote $y=0$ of $e^x$ becomes the asymptote $x=0$ of $\ln x$, and the point $(0,1)$ becomes $(1,0)$. The domain of one is the range of the other. Example: $\ln(e^{5})+e^{\ln5}=5+5=10$.

To undo a logarithm, raise $e$ to both sides: $\ln(ax+b)=q\ \Rightarrow\ ax+b=e^{q}\ \Rightarrow\ x=\frac{e^{q}-b}{a}.$ Unlike $e^{ax+b}=p$, a logarithm equation can be solved for any real $q$, including negative values. The solution is valid only if $ax+b>0$, which holds automatically when you substitute back $e^q>0$. Worked example: $\ln(2x+5)=3$ gives $2x+5=e^{3}$, so $x=\frac{e^{3}-5}{2}=7.54$ (3 s.f.). For $\ln(3x-2)=-2$: $3x-2=e^{-2}$ and $x=\frac{2+e^{-2}}{3}$.

For $y=\ln(ax+b)$, the vertical asymptote is where $ax+b=0$, i.e. $x=-\frac ba$. The curve crosses the $y$-axis at $(0,\ln b)$ (if $b>0$) and crosses the $x$-axis where $ax+b=1$. A graph such as $y=\ln(2x+5)$ is $y=\ln x$ after a translation and horizontal stretch: asymptote $x=-\frac52$, $x$-intercept $(-2,0)$, $y$-intercept $(0,\ln5)$. In models such as $H=2\ln(3t+1)+0.5$, the log increases without bound but ever more slowly. To find a time for a given value, isolate the $\ln$ and apply $e^{(\ldots)}$. Comment: the model has no upper limit, so it suits the early stage of growth only.