2.9 Exponential and logarithmic functionsIB Maths: Analysis and Approaches SL: Revision notes

An exponential function has the variable in the exponent: $f(x) = a^{x}$ with $a > 0$, $a \ne 1$. The most important is $f(x) = e^{x}$, where $e \approx 2.718$.

Key features of $y = a^{x}$:

• Passes through $(0, 1)$, since $a^{0} = 1$.
• Horizontal asymptote $y = 0$; the graph never touches the $x$-axis.
• Domain $x \in \mathbb{R}$, range $y > 0$.
• If $a > 1$ the function is increasing (growth); if $0 < a < 1$ it is decreasing (decay).

$2^{-x} = \left(\frac{1}{2}\right)^{x}$ is a decay function.

A logarithmic function is $f(x) = \log_{a} x$, $x > 0$. The natural logarithm is $\ln x = \log_{e} x$.

$\log_{a} x$ answers: 'what power of $a$ gives $x$?' So $\log_{2} 8 = 3$ and $\log_{3} \frac{1}{9} = -2$.

Key features of $y = \log_{a} x$ (for $a > 1$):

• Passes through $(1, 0)$, since $\log_{a} 1 = 0$.
• Vertical asymptote $x = 0$.
• Domain $x > 0$, range $y \in \mathbb{R}$.
• Increasing, but more and more slowly.

$y = a^{x}$ and $y = \log_{a} x$ are inverse functions: $\log_{a} a^{x} = x \quad (x \in \mathbb{R}), \qquad a^{\log_{a} x} = x \quad (x > 0).$ So $\ln e^{3} = 3$, $e^{\ln 7} = 7$ and $3^{\log_{3} 5} = 5$.

Graphically, $y = \log_{a} x$ is the reflection of $y = a^{x}$ in the line $y = x$: the point $(0, 1)$ becomes $(1, 0)$, the horizontal asymptote $y = 0$ becomes the vertical asymptote $x = 0$, and the range $y > 0$ becomes the domain $x > 0$.

Any exponential can be written with base $e$: $a^{x} = e^{x \ln a}.$ This works because $a = e^{\ln a}$, so $a^{x} = (e^{\ln a})^{x} = e^{x \ln a}$.

Example: a half-life model $m = 80 \times 2^{-\frac{t}{12}}$ becomes $m = 80e^{-kt}$ with $k = \frac{\ln 2}{12} \approx 0.0578$. Base-$e$ forms are standard in modelling and in calculus.

For growth or decay, $f(t) = A a^{t}$ or $f(t) = Ae^{kt}$: $A$ is the initial value ($t = 0$).

• $k > 0$: growth; $k < 0$: decay.
• A horizontal asymptote gives the long-term behaviour: in a decay model the quantity approaches 0 (or another value, such as room temperature in a cooling model) but never reaches it.

For $m(t) = 80 \times 2^{-\frac{t}{12}}$, the mass halves every 12 days: 80, 40, 20, 10 g at $t = 0, 12, 24, 36$.