Exponential and logarithmic functions (TCE Mathematics Methods, Tasmania)

What this dot point is asking

Exponential and logarithmic functions describe anything that grows or shrinks at a rate proportional to its current size: populations, money at compound interest, radioactive decay, cooling. This dot point develops the algebra, the calculus, and the modelling.

Functions and their inverses

The exponential function $y = a^x$ (with $a > 0$, $a \ne 1$) and the logarithm $y = \log_a x$ are inverses: $a^x = y \iff x = \log_a y$. Their graphs are reflections in the line $y = x$. The exponential has a horizontal asymptote $y = 0$; the logarithm has a vertical asymptote $x = 0$ and is only defined for $x > 0$.

Index and logarithm laws

The index laws ($a^m a^n = a^{m+n}$, $\dfrac{a^m}{a^n} = a^{m-n}$, $(a^m)^n = a^{mn}$) translate directly into log laws:

The change-of-base rule $\log_a x = \dfrac{\ln x}{\ln a}$ lets you evaluate any logarithm with the natural log on a calculator.

Derivatives

The standard derivatives are:

With the chain rule these generalise:

Solving exponential and logarithmic equations

Two standard techniques solve almost every equation in this dot point.

To solve an equation where the unknown sits in an exponent, isolate the exponential and take the natural log of both sides. For example, $3\times 2^{x} = 96$ gives $2^{x} = 32$, then $x\ln 2 = \ln 32$, so $x = \dfrac{\ln 32}{\ln 2} = 5$. The logarithm "brings the power down" because $\ln(a^{x}) = x\ln a$.

To solve a logarithmic equation, combine every logarithm into a single one using the laws, then either equate arguments (if both sides are logs of the same base) or rewrite in exponential form. For instance, $\log_{2}(x) + \log_{2}(x - 2) = 3$ becomes $\log_{2}\big(x(x - 2)\big) = 3$, so $x(x - 2) = 2^{3} = 8$, giving $x^{2} - 2x - 8 = 0$ and $x = 4$ (rejecting $x = -2$, which makes the argument negative).

The change-of-base rule in practice

Calculators evaluate $\ln$ and $\log_{10}$, but exam questions can use any base. The change-of-base rule $\log_{a}x = \dfrac{\ln x}{\ln a}$ converts to a base you can evaluate. For example, $\log_{5}40 = \dfrac{\ln 40}{\ln 5} = \dfrac{3.6889}{1.6094} \approx 2.292$. The same rule explains why $\dfrac{d}{dx}\log_{a}x = \dfrac{1}{x\ln a}$: writing $\log_{a}x = \dfrac{\ln x}{\ln a}$ and differentiating the constant multiple $\dfrac{1}{\ln a}$ of $\ln x$ gives the result directly.

Growth and decay models

Exponential models take the form $A(t) = A_0 e^{kt}$. A positive $k$ is growth; a negative $k$ is decay. The constant $A_0$ is the initial amount (at $t = 0$). For radioactive decay the half-life $T$ satisfies $e^{kT} = \tfrac12$, giving $k = \dfrac{\ln(1/2)}{T}$. Always read whether the question gives you a rate, a doubling time, or a half-life, and convert it to $k$ first.

Why these models recur

The defining property $\dfrac{dA}{dt} = kA$ means the rate of change is proportional to the current amount, which is exactly the situation for unchecked population growth, compound interest, radioactive decay, the cooling of an object (Newton's law of cooling), and the elimination of a drug from the bloodstream. Recognising this signature lets you write down the model immediately and read $k$ from the exponent.