The logarithmic function base e
The natural logarithm ln x as the inverse of e^x, combining ln and exp, solving equations with ln, and transformations of y = ln x.
The natural logarithm
All the log laws hold: $\ln xy = \ln x + \ln y$, $\ln \dfrac{x}{y} = \ln x - \ln y$, $\ln x^n = n\ln x$, and $\ln 1 = 0$, $\ln e = 1$.
Orange: y = eˣ. Blue: y = ln x, its reflection in y = x. Move P along the log curve.
Where is $\ln x$ negative?
Solving equations
Solve $e^x = 5$ and $\ln x = 2$, correct to 3 decimal places.
Solve $e^{2x} = 9$ exactly.
Transformations of y = ln x
$y = \ln x$ passes through $(1, 0)$ with gradient $1$ there, and has vertical asymptote $x = 0$. $y = \ln(x - h) + k$ has asymptote $x = h$ and domain $x > h$. $y = \ln(-x)$ reflects in the $y$-axis and has domain $x < 0$.
State the domain of $y = \ln(x - 3)$ and find its $x$-intercept.
$\ln x$ means:
The natural logarithm uses base $e$.
Simplify $\ln e^3$.
$\ln$ undoes $e$.
Simplify $e^{\ln 7}$.
$e$ undoes $\ln$.
Solve $e^x = 5$, correct to 3 decimal places.
$x = \ln 5$.
Solve $\ln x = 2$, correct to 3 decimal places.
$x = e^2$.
Evaluate $\ln 1$.
$e^0 = 1$.
The domain of $y = \ln(x - 3)$ is:
The log needs a positive argument.
Solve $e^{2x} = 9$, correct to 3 decimal places.
$x = \tfrac12\ln 9 = \ln 3$.
Simplify $\ln\left(e^2 \times e^3\right)$.
$\ln e^5 = 5$.
New numbers every time. Keep going until you get five in a row.
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