Year 12 Methods (Unit 3 & 4) Exponential and logarithmic functions

Revision of exponential functions

Q: What is an exponential function?

An exponential function has the form y equals r to the power x, where the base r is a fixed positive number and the variable x is the exponent. Every such graph passes through (0, 1), has the horizontal asymptote y equals 0, a domain of all real numbers and a range of y greater than 0.

Q: What is the horizontal asymptote of y = r^x?

For the basic exponential y equals r to the power x the horizontal asymptote is the line y equals 0 (the x-axis). The curve approaches it but never touches it. After a vertical shift, y equals r to the power (x minus h) plus k, the asymptote moves to y equals k.

Q: What does y = r^(x-h)+k do to the graph of y = r^x?

The value h shifts the graph horizontally (right when h is positive, left when h is negative) and k shifts it vertically. The k also moves the horizontal asymptote to y equals k, so the range becomes y greater than k (or y less than k if the graph is reflected in the x-axis).

Q: When is an exponential function growth or decay?

It depends on the base. When r is greater than 1 the function increases - exponential growth. When r is between 0 and 1 the function decreases - exponential decay. In the model N equals N0 times r to the power t, r greater than 1 grows and 0 less than r less than 1 decays.

Q: What does a factor of 0.85 mean in an exponential model?

Multiplying by 0.85 each period means the quantity keeps 85 percent of its value, which is a 15 percent decrease per period (decay). A factor of 1.05 means 5 percent growth per period. The percentage change is the factor compared with 1.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), the exponential function $y=r^x$ (with base $r>0$) passes through $(0,1)$, has the horizontal asymptote $y=0$, and either grows ($r>1$) or decays ($0<r<1$) — revised here alongside its transformations $y=r^{x-h}+k$, solving equations by equating indices, and growth-and-decay models, ready for the calculus of Unit 3.

An exponential function has the form $y=r^x$, where the base $r$ is a fixed positive number ($r>0$, $r\neq 1$) and the variable $x$ sits in the exponent. Because $r^0=1$, every such graph passes through the $y$-intercept $(0,1)$.

As $x$ moves in one direction the curve races upward, and in the other it flattens towards the $x$-axis without ever reaching it — the line $y=0$ is a horizontal asymptote. The domain is all real $x$ and the range is $y>0$.

The base decides the shape: when $r>1$ the function is increasing (exponential growth); when $0<r<1$ it is decreasing (exponential decay). Shifting the graph with $y=r^{x-h}+k$ moves the asymptote to $y=k$. (The natural base $e$ and its calculus come later, in Unit 3.)

Key idea. \(y=r^x\) (\(r>0\)): \(y\)-intercept \((0,1)\), asymptote \(y=0\), range \(y>0\). Increasing when \(r>1\), decreasing when \(0<r<1\).

$y=2^x$ ($r>1$, growth) and $y=\left(\dfrac12\right)^x$ ($0<r<1$, decay): both through $(0,1)$, asymptote $y=0$

$y=2^{x+2}-1$: shifted left and down, asymptote $y=-1$, $y$-intercept $(0,3)$

The basic exponential function (base $r>0$):

\[y=r^x \qquad r>0\]

y = r^{x}

The transformed exponential (horizontal shift $h$, vertical shift $k$):

\[y=r^{x-h}+k \qquad \text{asymptote } y=k\]

y = r^{x - h} + k

The growth-and-decay model over time $t$, with initial amount $N_0$:

\[N=N_0\,r^{t} \qquad r>1:\ \text{growth} \quad 0<r<1:\ \text{decay}\]

N = N_{0} r^{t}

Reading the base. A factor \(r=0.85\) keeps \(85\%\) each period — a \(15\%\) decrease; a factor \(r=1.05\) is \(5\%\) growth. Compare the base with \(1\).

How to sketch $y=r^{x-h}+k$

Asymptote first. The $+k$ lifts the whole curve, so the horizontal asymptote is $y=k$ (not $y=0$).
Find the $y$-intercept. Substitute $x=0$ into $y=r^{x-h}+k$ and simplify.
Set the shape. Increasing when $r>1$, decreasing when $0<r<1$; the range is $y>k$.

Solving exponential equations. Write both sides as a power of the same base, then equate the indices — e.g. \(2^x=32=2^5\) gives \(x=5\). If equal bases are not possible, solve with technology.

Example 1 — Features of $y=r^x$

State the $y$-intercept, horizontal asymptote, range, and whether $y=\left(\dfrac14\right)^x$ is increasing or decreasing.

Solution

$y$-intercept: at $x=0$, $\left(\dfrac14\right)^0=1$, so $(0,1)$.

Asymptote: $y=0$; range: $y>0$.

Here $r=\dfrac14$, and $0<r<1$, so the graph is decreasing (decay).

{(\frac{1}{4})}^{0} = 1

Example 2 — Transformation

For $y=2^{x+2}-1$, state the horizontal asymptote, the $y$-intercept and the range.

Solution

Write it as $y=2^{x-(-2)}+(-1)$: shifted left $2$ and down $1$.

$\text{asymptote}$	$:$	$y=-1$
$y(0)$	$=$	$2^{2}-1=3$
$\text{range}$	$:$	$y>-1$

Asymptote $y=-1$, $y$-intercept $(0,3)$, range $y>-1$.

y = 2^{2} - 1 = 3

Example 3 — Equate the indices

Solve for $x$: (a) $5^x=125$; (b) $2^x=\dfrac{1}{16}$.

Solution

(a) Write $125$ as a power of $5$.

$5^x$	$=$	$5^3$
$x$	$=$	$3$

(b) Write $\dfrac{1}{16}=2^{-4}$.

$2^x$	$=$	$2^{-4}$
$x$	$=$	$-4$

5^{x} = 5^{3}, x = 3

Example 4 — Decay model

A car bought for $\$36\,000$ loses $20\%$ of its value each year. Find its value after $3$ years.

Solution

Losing $20\%$ keeps $80\%$, so the factor is $r=0.8$ (decay).

$V$	$=$	$36\,000\times 0.8^{\,t}$
$V(3)$	$=$	$36\,000\times 0.8^{3}$
$V(3)$	$=$	$36\,000\times 0.512$
$V(3)$	$=$	$\$18\,432$

After $3$ years the car is worth $\$18\,432$.

36000 \times {0.8}^{3} = 18432

Common pitfalls

Every $y=r^x$ passes through $(0,1)$. Because $r^0=1$ for any base, changing the base makes the curve steeper or flatter but does not move the $y$-intercept away from $(0,1)$.

The asymptote moves with $k$, not $h$. For $y=r^{x-h}+k$ the horizontal asymptote is $y=k$ (the vertical shift), so the range is $y>k$. The $h$ only slides the graph sideways.

Growth vs decay is about the base. $0<r<1$ gives decay, $r>1$ gives growth. A factor of $0.85$ is a $15\%$ decrease per period — not $85\%$.

Frequently asked questions

What is an exponential function?

It has the form $y=r^x$, where the base $r$ is a fixed positive number and the variable $x$ is the exponent. Every such graph passes through $(0,1)$, has asymptote $y=0$, domain all real $x$ and range $y>0$.

What is the horizontal asymptote of y = r^x?

The line $y=0$ (the $x$-axis): the curve approaches it but never touches it. After a vertical shift $y=r^{x-h}+k$ the asymptote moves to $y=k$.

How do you solve an exponential equation with the same base?

Write both sides as a power of the same base, then equate the indices. For example $2^x=32=2^5$ gives $x=5$. If equal bases are not possible, solve with technology.

What does y = r^(x-h)+k do to the graph of y = r^x?

$h$ shifts it horizontally (right if $h>0$, left if $h<0$) and $k$ shifts it vertically, moving the asymptote to $y=k$ so the range becomes $y>k$.

When is an exponential function growth or decay?

When $r>1$ it increases (growth); when $0<r<1$ it decreases (decay). In $N=N_0\,r^{t}$ the base $r$ decides which.

What does a factor of 0.85 mean in an exponential model?

Multiplying by $0.85$ keeps $85\%$ each period, a $15\%$ decrease (decay). A factor of $1.05$ is $5\%$ growth. Compare the base with $1$.

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More in Exponential and logarithmic functions

\(\text{asymptote}\)	\(:\)	\(y=-1\)
\(y(0)\)	\(=\)	\(2^{2}-1=3\)
\(\text{range}\)	\(:\)	\(y>-1\)

\(5^x\)	\(=\)	\(5^3\)
\(x\)	\(=\)	\(3\)

\(2^x\)	\(=\)	\(2^{-4}\)
\(x\)	\(=\)	\(-4\)

\(V\)	\(=\)	\(36\,000\times 0.8^{\,t}\)
\(V(3)\)	\(=\)	\(36\,000\times 0.8^{3}\)
\(V(3)\)	\(=\)	\(36\,000\times 0.512\)
\(V(3)\)	\(=\)	\(\$18\,432\)

Revision of exponential functions

How to sketch \(y=r^{x-h}+k\)

Common pitfalls

Frequently asked questions