How are the graphs of y=e^x and y=\ln x related?

They are reflections of each other in y=x.

ax+b=e^{q}, so x=\frac{e^{q}-b}{a}.

Why can \ln x=-3 be solved but e^{x}=-3 cannot?

\ln x takes all real values, but e^x is always positive.

The natural logarithm ln xEdexcel International A Level Maths: Flashcards

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Question

State the domain and range of $y=\ln x$.

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State the domain and range of $y=\ln x$ .: Domain $x>0$ ; range all real numbers.
Where does $y=\ln x$ cross the $x$ -axis?: At $(1,0)$ .
What is the asymptote of $y=\ln x$ ?: The $y$ -axis, $x=0$ .
Simplify $\ln(e^{x})$ .: $x$ .
Simplify $e^{\ln x}$ .: $x$ , for $x>0$ .
What is $\ln e$ ?: $1$ .
What is $\ln 1$ ?: $0$ .
How are the graphs of $y=e^x$ and $y=\ln x$ related?: They are reflections of each other in $y=x$ .
Solve $\ln(ax+b)=q$ .: $ax+b=e^{q}$ , so $x=\frac{e^{q}-b}{a}$ .
What is the domain of $\ln(2x+5)$ ?: $x>-\frac52$ .
What is the vertical asymptote of $y=\ln(ax+b)$ ?: $x=-\frac{b}{a}$ .
Why can $\ln x=-3$ be solved but $e^{x}=-3$ cannot?: $\ln x$ takes all real values, but $e^x$ is always positive.
Solve $\ln(x-1)=0$ .: $x-1=e^{0}=1$ , so $x=2$ .

Exam questions on The natural logarithm ln x

The function $f$ is defined by $f(x)=\ln(3x-2)$ , where $x$ takes the largest possible domain.
Find the exact solution of $f(x)=-2$ .2 marks
The functions $g$ and $h$ are defined by $g(x)=e^{x}$ and $h(x)=\ln x$ , for $x>0$ in the case of $h$ .
State the equation of the asymptote of the graph of $y=h(x)$ and the coordinates of the point where it crosses the $x$ -axis.2 marks
The curve $C$ has equation $y=\ln(2x+5)$ .
Show that the solution of $\ln(2x+5)=3$ is $x=\frac{e^{3}-5}{2}$ and find its value to 3 significant figures.3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).