Exponential and Logarithmic Functions: Year 11 Guide

At \(x=0\), the exponential curves \(y=2^x\), \(y=3^x\), and \(y=(\frac12)^x\) all pass through exactly the same point: \((0,1)\). Yet one rises faster than the other, while the third decreases. What controls these differences? And is there a particular exponential curve whose tangent gradient at \((0,1)\) is exactly 1?

These questions lead to two important ideas in Year 11 Mathematics Advanced: Euler’s number \(e\), and logarithms, which allow us to work backwards and find an unknown exponent.

This guide covers the Exponential and logarithmic functions focus area of the NSW Mathematics Advanced 11-12 Syllabus (2024). You will learn how to graph exponential and logarithmic functions, understand the significance of \(e\), apply logarithmic laws, solve exponential equations, and connect exponential and logarithmic graphs through reflections.

01Understanding exponential functions

An exponential function is a function in which the variable appears in the exponent.

Consider:

\[ y=2^x \]

You can evaluate this function by substituting different values of \(x\).

Notice the pattern. Every time \(x\) increases by 1, the output doubles.

This is what makes the function exponential. Instead of adding a constant amount each time, we multiply by a constant factor.

1.1 The general exponential functions

The NSW syllabus considers two important families of exponential functions:

\[ y=k(a^x) \]

and

\[ y=k(a^{-x}) \]

where:

• \(a\) is the base, with \(a>0\) and \(a\ne1\).
• \(k\) is a non-zero constant, so \(k\ne0\).
• \(x\) is any real number.

The base \(a\) determines whether the exponential part increases or decreases. The constant \(k\) controls the vertical stretch and whether the graph is reflected across the \(x\)-axis.

Let’s examine these effects separately.

When \(a>1\)

The function \(y=a^x\) is increasing.

For example, \(2^x\) doubles whenever \(x\) increases by 1.

The function \(y=a^{-x}\), however, is decreasing. This is because:

\[ a^{-x}=\left(\frac{1}{a}\right)^x \]

When \(a>1\), its reciprocal is between 0 and 1.

For example:

\[ 2^{-x}=\left(\frac12\right)^x \]

Each increase of 1 in \(x\) halves the output.

When \(0<a<1\)

The behaviour reverses.

The function \(y=a^x\) is decreasing because multiplying repeatedly by a fraction between 0 and 1 makes the output smaller.

Meanwhile, \(y=a^{-x}\) is increasing because its effective base, \(1/a\), is greater than 1.

The four possibilities are summarised below.

What happens if \(k\) is negative?

Multiplying a function by a negative number reflects its graph across the \(x\)-axis. Its increasing or decreasing behaviour reverses.

For example, \(y=-2^x\) is decreasing, even though \(y=2^x\) is increasing.

1.2 Domain, range, intercepts and asymptotes

These characteristics are particularly important when sketching exponential functions.

Consider:

\[ y=k(a^x) \]

Since \(a\) is positive, \(a^x\) is positive for every real value of \(x\).

Multiplying by \(k\) does not change the fact that the function is defined for every real \(x\).

Therefore, the domain is:

\[ \boxed{x\in\mathbb R} \]

Now consider the \(y\)-intercept.

At the \(y\)-intercept, \(x=0\). Substituting:

\[ \begin{aligned} y&=k(a^0)\\ &=k(1)\\ &=k \end{aligned} \]

Therefore, the \(y\)-intercept is:

\[ \boxed{(0,k)} \]

The same result applies to \(y=k(a^{-x})\), because \(a^0=1\).

Next, consider the range.

• If \(k>0\), the function is always positive, so its range is \(y>0\).
• If \(k<0\), the function is always negative, so its range is \(y<0\).

Finally, consider the horizontal asymptote.

As the exponential part approaches zero, the function approaches:

\[ \boxed{y=0} \]

The graph never actually reaches this line because an exponential expression with a positive base is never zero.

This gives us a useful summary.

1.3 End behaviour of exponential functions

End behaviour describes what happens to a function as \(x\) becomes extremely large or extremely negative.

For exponential functions, the direction of growth depends on the base.

For \(a>1\):

• \(a^x\) increases without bound as \(x\to\infty\).
• \(a^x\) approaches zero as \(x\to-\infty\).
• \(a^{-x}\) approaches zero as \(x\to\infty\).
• \(a^{-x}\) increases without bound as \(x\to-\infty\).

For \(0<a<1\), these behaviours reverse.

The complete summary is:

In this table, an unbounded function approaches \(+\infty\) when \(k>0\), and \(-\infty\) when \(k<0\).

Similarly, an approach to zero is from above when \(k>0\), and from below when \(k<0\).

02Euler’s number and the exponential function

Remember the opening question about the gradients of exponential curves?

All functions of the form \(y=a^x\) pass through \((0,1)\). However, they do not have the same tangent gradient at that point.

This is where a special mathematical constant appears.

2.1 Investigating the gradient at the \(y\)-intercept

The gradient of a tangent tells us how quickly a curve is changing at a particular point.

For \(y=a^x\), we can use a graphing application to draw the curve and its tangent at \(x=0\).

Now vary the base \(a\).

The following approximate values show how the tangent gradient changes.

Notice three things.

First, when \(0<a<1\), the gradient is negative because the exponential curve is decreasing.

Second, when \(a>1\), the gradient is positive because the curve is increasing.

Third, there is a particular base for which the tangent gradient is exactly 1.

Graphing applications allow us to investigate this relationship by varying \(a\) and measuring the gradient of the tangent at \((0,1)\).

The investigation identifies one unique base for which the tangent gradient is 1.

This number is:

\[ \boxed{e\approx2.71828182845\ldots} \]

It is called Euler’s number.

2.2 Why does the derivative of an exponential function have the same shape?

The next syllabus requirement is to investigate the gradient function of:

\[ y=a^x \]

The gradient function, also called the derivative, tells us the tangent gradient at every point on the curve.

Let’s see why it has a particularly simple form.

Consider a small horizontal change \(h\).

The corresponding change in the function is:

\[ a^{x+h}-a^x \]

Using the laws of indices:

\[ a^{x+h}=a^x a^h \]

Therefore:

\[ a^{x+h}-a^x=a^x(a^h-1) \]

To calculate the gradient, divide by the horizontal change:

\[ \frac{a^{x+h}-a^x}{h} = a^x\left(\frac{a^h-1}{h}\right) \]

As \(h\) approaches zero, the expression in brackets approaches a constant that depends only on the base \(a\).

Call this constant \(c_a\).

The derivative therefore has the form:

\[ \boxed{\frac{d}{dx}(a^x)=c_a a^x} \]

The important observation is that the derivative is proportional to the original exponential function.

The constant \(c_a\) depends on the base, not on \(x\).

2.3 Why is \(e^x\) its own derivative?

We now have two important facts:

• The derivative of \(a^x\) is \(c_a a^x\).
• There is one unique base \(e\) for which the tangent gradient at \(x=0\) is 1.

At \(x=0\), the exponential function has value 1. Therefore, its derivative at this point is \(c_a\).

For the special base \(e\), we have:

\[ c_e=1 \]

Substituting this into the general derivative result gives:

\[ \frac{d}{dx}(e^x)=1(e^x) \]

Hence:

\[ \boxed{\frac{d}{dx}(e^x)=e^x} \]

This is one of the most important properties of Euler’s number.

The exponential function \(e^x\) is its own derivative.

In other words, at every point on the curve, the tangent gradient is equal to the \(y\)-coordinate of that point.

For example, at \(x=0\), the function has value 1 and gradient 1. At \(x=2\), both its value and tangent gradient are \(e^2\).

This property makes \(e^x\) particularly important when studying differentiation, continuous growth, and exponential decay.

03Logarithms: working backwards from an exponent

Suppose you know that:

\[ 2^3=8 \]

Now imagine that you are given the number 8 and asked to find the power to which 2 must be raised to produce it.

You are looking for the exponent.

The answer is 3.

Logarithms provide a notation for finding this unknown exponent.

3.1 The definition of a logarithm

The logarithm of a positive number \(y\) to a positive base \(a\) is the index to which \(a\) must be raised to produce \(y\).

We write:

\[ \boxed{\log_a y=x} \]

if and only if:

\[ \boxed{a^x=y} \]

For example:

\[ 2^3=8 \]

is equivalent to:

\[ \log_2 8=3 \]

Both statements describe exactly the same relationship.

The logarithm simply makes the unknown exponent the subject.

The general equivalence is:

\[ \boxed{y=a^x\iff x=\log_a y} \]

The conditions are:

\[ a>0,\qquad a\ne1,\qquad y>0 \]

Why are these restrictions necessary?

An exponential function with a positive base always produces a positive output. Therefore, there is no real exponent that makes a positive base produce zero or a negative number.

Also, a base of 1 is unsuitable because:

\[ 1^x=1 \]

for every real \(x\). It cannot generate other positive numbers.

A negative base is also unsuitable for a logarithmic function defined over all real exponents.

3.2 Understanding positive and negative logarithms

The exponent can be positive, negative, or zero.

For example:

\[ \log_2 8=3 \]

because:

\[ 2^3=8 \]

But:

\[ \log_2\left(\frac18\right)=-3 \]

because:

\[ 2^{-3}=\frac18 \]

And:

\[ \log_2 1=0 \]

because:

\[ 2^0=1 \]

A logarithm is not restricted to positive answers. Its argument must be positive, but its output can be any real number.

3.3 Solving exponential equations using logarithms

Consider an equation of the form:

\[ a^x=b \]

If \(a>0\), \(a\ne1\), and \(b>0\), we can use the logarithmic definition directly:

\[ \boxed{x=\log_a b} \]

For example:

\[ 3^x=20 \]

Since 20 is not an obvious integer power of 3:

\[ x=\log_3 20 \]

A calculator gives:

\[ \boxed{x\approx2.727} \]

This is the main reason logarithms are useful when solving exponential equations. They allow us to find an exponent even when the answer is not a simple integer.

If \(b\leq0\), there is no real solution because a positive base raised to a real power cannot produce a non-positive result.

If \(a=1\), the equation must be considered separately because its left-hand side is always 1.

3.4 The natural logarithm

We have already introduced Euler’s number:

\[ e\approx2.71828182845\ldots \]

A logarithm with base \(e\) has a special name: the natural logarithm.

Its notation is:

\[ \boxed{\ln x=\log_e x} \]

For example:

\[ \ln(e^4)=4 \]

The natural logarithm is the inverse function of \(e^x\).

It is particularly useful because \(e^x\) has the special derivative property introduced earlier.

3.5 The inverse properties of logarithms and exponentials

Since logarithms and exponentials undo each other, we have two important identities.

The first is:

\[ \boxed{\log_a(a^x)=x} \]

This is true for every real \(x\).

For example:

\[ \log_3(3^7)=7 \]

The second is:

\[ \boxed{a^{\log_a x}=x} \]

This is true when \(x>0\).

For example:

\[ 5^{\log_5 12}=12 \]

Notice that the order matters.

In the first identity, the logarithm removes the exponential.

In the second identity, the exponential removes the logarithm.

Both identities rely on using the same base.

04The laws of logarithms

Logarithmic laws allow us to simplify expressions and solve equations without having to calculate every logarithm separately.

Rather than memorising the laws without understanding them, let’s derive them from the index laws you already know.

Throughout this section, assume:

\[ a>0,\qquad a\ne1 \]

and all logarithmic arguments are positive.

4.1 The product law

We want to simplify:

\[ \log_a m+\log_a n \]

Let:

\[ p=\log_a m \]

and:

\[ q=\log_a n \]

Using the definition of a logarithm:

\[ m=a^p,\qquad n=a^q \]

Now multiply these expressions:

\[ mn=a^p a^q \]

Using the index law:

\[ mn=a^{p+q} \]

Taking the logarithm to base \(a\):

\[ \log_a(mn)=p+q \]

Substituting our original definitions gives:

\[ \boxed{\log_a m+\log_a n=\log_a(mn)} \]

The product of two logarithmic arguments becomes the sum of their logarithms.

For example:

\[ \log_2 4+\log_2 8 \]

can be combined into:

\[ \log_2(4\times8)=\log_2 32=5 \]

4.2 The quotient law

Now consider:

\[ \log_a m-\log_a n \]

Using the same definitions:

\[ m=a^p,\qquad n=a^q \]

Divide the two expressions:

\[ \frac mn=\frac{a^p}{a^q} \]

Using the index law for division:

\[ \frac mn=a^{p-q} \]

Taking logarithms:

\[ \log_a\left(\frac mn\right)=p-q \]

Therefore:

\[ \boxed{\log_a m-\log_a n =\log_a\left(\frac mn\right)} \]

For example:

\[ \log_3 81-\log_3 9 \]

becomes:

\[ \log_3\left(\frac{81}{9}\right) =\log_3 9 =2 \]

4.3 The power law

Consider:

\[ \log_a(m^n) \]

Let:

\[ p=\log_a m \]

Therefore:

\[ m=a^p \]

Raise both sides to the power \(n\):

\[ m^n=(a^p)^n \]

Using the index law:

\[ m^n=a^{np} \]

Taking logarithms:

\[ \log_a(m^n)=np \]

Substituting \(p=\log_a m\):

\[ \boxed{\log_a(m^n)=n\log_a m} \]

For example:

\[ \log_2(8^3) \]

can be simplified to:

\[ 3\log_2 8=3(3)=9 \]

These three laws are the foundation of logarithmic algebra.

4.4 Three important logarithmic results

Some logarithmic expressions appear so frequently that they are worth recognising immediately.

Result 1: The logarithm of the base is 1.

\[ \boxed{\log_a a=1} \]

This follows because:

\[ a^1=a \]

Result 2: The logarithm of 1 is 0.

\[ \boxed{\log_a1=0} \]

This follows because:

\[ a^0=1 \]

Result 3: The logarithm of a reciprocal is the negative logarithm.

For \(x>0\):

\[ \begin{aligned} \log_a\left(\frac1x\right) &=\log_a(x^{-1})\\ &=-\log_a x \end{aligned} \]

Therefore:

\[ \boxed{\log_a\left(\frac1x\right)=-\log_a x} \]

This result is particularly useful when simplifying expressions containing fractions.

4.5 A common logarithmic mistake

A tempting mistake is to assume that logarithms distribute over addition.

For example, a student might write:

\[ \log_a(m+n)=\log_a m+\log_a n \]

This is incorrect.

The product law only works when the arguments are multiplied.

A quick counterexample makes this clear:

\[ \log_2(1+1)=\log_2 2=1 \]

But:

\[ \log_2 1+\log_2 1=0+0=0 \]

The results are different.

Remember:

\[ \boxed{\log_a(mn)=\log_a m+\log_a n} \]

There is no equivalent addition law.

05The change of base rule

What happens when you need to calculate:

\[ \log_5 17 \]

but your calculator only has buttons for \(\log\) and \(\ln\)?

You need the change of base rule.

This rule allows you to rewrite a logarithm using a different base.

5.1 Deriving the change of base rule

Suppose:

\[ t=\log_a x \]

Using the definition of a logarithm:

\[ a^t=x \]

Now take the logarithm of both sides using a new base \(b\):

\[ \log_b(a^t)=\log_b x \]

Apply the power law:

\[ t\log_b a=\log_b x \]

Rearranging:

\[ t=\frac{\log_b x}{\log_b a} \]

Since \(t=\log_a x\), we obtain:

\[ \boxed{\log_a x=\frac{\log_b x}{\log_b a}} \]

The conditions are:

\[ a>0,\quad a\ne1,\quad b>0,\quad b\ne1,\quad x>0 \]

This formula works for any valid new base \(b\).

5.2 Using natural logarithms

The most common application is choosing \(b=e\).

Since:

\[ \log_e x=\ln x \]

the change of base rule becomes:

\[ \boxed{\log_a x=\frac{\ln x}{\ln a}} \]

You can also use common logarithms:

\[ \log_a x=\frac{\log x}{\log a} \]

where \(\log\) represents the base-10 logarithm.

Both methods give the same result.

06Using digital tools with exponential and logarithmic expressions

The NSW syllabus expects students to use digital tools when investigating exponential and logarithmic expressions.

A scientific calculator or graphing application can help you:

• Evaluate exponential expressions.
• Calculate logarithms with different bases.
• Approximate irrational values.
• Investigate tangent gradients.
• Compare exponential and logarithmic graphs.
• Check algebraic solutions.

However, there is an important distinction between finding a decimal approximation and establishing whether an exact value is rational or irrational.

6.1 Rational and irrational values

A rational number can be written as a fraction of two integers.

For example:

\[ \frac34,\quad 5,\quad -2.5 \]

An irrational number cannot be written as a ratio of two integers.

Its decimal expansion neither terminates nor repeats.

Consider these examples.

Let’s examine the last two examples more carefully.

For:

\[ 16^{3/4} \]

we can write:

\[ 16^{3/4}=(\sqrt[4]{16})^3 \]

Therefore:

\[ 16^{3/4}=2^3=8 \]

The value is rational.

For:

\[ 2^{1/2}=\sqrt2 \]

a calculator gives:

\[ \sqrt2\approx1.41421356 \]

This value is irrational.

A decimal approximation suggests its size, but the fact that \(\sqrt2\) is irrational comes from mathematical reasoning, not the calculator display.

Now consider:

\[ \log_2 3 \]

A calculator gives:

\[ \log_2 3\approx1.5849625 \]

We can actually prove that this value is irrational.

Suppose it were rational. Then we could write:

\[ \log_2 3=\frac pq \]

where \(p\) and \(q\) are positive integers.

Using the logarithmic definition:

\[ 2^{p/q}=3 \]

Raise both sides to the power \(q\):

\[ 2^p=3^q \]

But the left-hand side contains only the prime factor 2, while the right-hand side contains only the prime factor 3.

This is impossible for positive integers \(p\) and \(q\).

Therefore:

\[ \boxed{\log_2 3\text{ is irrational}} \]

The important distinction is that digital tools help you investigate and approximate values, while exact mathematical reasoning establishes their properties.

07Graphing logarithmic functions

We have now defined logarithms algebraically. It is time to examine their graphs.

Consider the general logarithmic function:

\[ \boxed{y=\log_a x} \]

where:

\[ a>0,\qquad a\ne1 \]

The behaviour of this graph depends on whether the base is greater than 1 or between 0 and 1.

7.1 The graph when \(a>1\)

Consider:

\[ y=\log_2 x \]

We can find some points by rewriting the logarithmic expression in exponential form.

For example:

\[ \log_2 8=3 \]

so the graph passes through \((8,3)\).

Similarly:

\[ \log_2 4=2 \]

and:

\[ \log_2 1=0 \]

This gives us the following points:

The graph is increasing because a larger positive input requires a larger exponent to produce it when the base is greater than 1.

Its key characteristics are:

• Domain: \(x>0\).
• Range: all real numbers.
• \(x\)-intercept: \((1,0)\).
• Vertical asymptote: \(x=0\).
• No \(y\)-intercept.

Why is there no \(y\)-intercept?

A \(y\)-intercept would require substituting \(x=0\). However, logarithms are only defined for positive arguments.

The graph approaches negative infinity as \(x\to0^+\), and increases without bound as \(x\to\infty\).

7.2 The graph when \(0<a<1\)

Now consider:

\[ y=\log_{\frac12}x \]

The base is between 0 and 1, so the function is decreasing.

For example:

\[ \log_{\frac12}2=-1 \]

because:

\[ \left(\frac12\right)^{-1}=2 \]

Similarly:

\[ \log_{\frac12}\left(\frac12\right)=1 \]

The graph passes through:

\[ (1,0),\quad \left(\frac12,1\right),\quad (2,-1) \]

Its characteristics are:

• Domain: \(x>0\).
• Range: all real numbers.
• \(x\)-intercept: \((1,0)\).
• Vertical asymptote: \(x=0\).
• No \(y\)-intercept.
• The function is decreasing.

Its end behaviour is the reverse of the graph with a base greater than 1.

As \(x\to0^+\):

\[ \log_{\frac12}x\to\infty \]

As \(x\to\infty\):

\[ \log_{\frac12}x\to-\infty \]

The comparison is summarised below.

08The reflection relationship between exponential and logarithmic graphs

This is one of the most important connections in the focus area.

Consider:

\[ y=a^x \]

and:

\[ y=\log_a x \]

We already know that logarithms undo exponentials.

This means the two functions are inverses of each other.

The graphs of inverse functions are reflections of each other across the line:

\[ \boxed{y=x} \]

8.1 Why are the graphs reflections?

Consider a point on the exponential graph:

\[ y=a^x \]

Suppose the point is:

\[ (p,q) \]

This means:

\[ q=a^p \]

Using the definition of a logarithm:

\[ p=\log_a q \]

Therefore, the logarithmic graph contains the point:

\[ (q,p) \]

Notice that the coordinates have swapped.

For example, if the exponential graph contains \((2,4)\), the logarithmic graph contains \((4,2)\).

Swapping the coordinates of every point produces a reflection across the line \(y=x\).

8.2 Reflection when \(a>1\)

For \(a>1\), both functions are increasing.

For example:

\[ y=2^x \]

and:

\[ y=\log_2 x \]

The exponential graph has a horizontal asymptote at \(y=0\).

After reflection, this becomes the vertical asymptote \(x=0\) of the logarithmic graph.

Similarly, the exponential \(y\)-intercept \((0,1)\) becomes the logarithmic \(x\)-intercept \((1,0)\).

The domain and range also swap:

8.3 Reflection when \(0<a<1\)

The same reflection relationship applies when the base is between 0 and 1.

Consider:

\[ y=\left(\frac12\right)^x \]

and:

\[ y=\log_{\frac12}x \]

Both functions are decreasing.

Their graphs are still reflections across \(y=x\).

This is important because students sometimes assume that inverse functions must always be increasing. That is incorrect.

The direction of the graph depends on the base, but the inverse relationship remains unchanged.

When using a graphing application, try plotting both functions for \(a=2\), then repeat with \(a=\frac12\). Add the line \(y=x\) to each graph.

You should see the reflection relationship in both cases.

09Questions and solutions

The following questions progress from straightforward applications of logarithmic laws to more challenging problems involving inverse functions, graph behaviour, and mathematical reasoning.

Question 1

Evaluate each expression exactly.

(a) \(\log_3 81\)

(b) \(5^{\log_5 7}\)

(c) \(\log_2\left(\frac18\right)\)

Show the exponential or logarithmic property used in each case.

Solution 1

The answers are:

\[ \boxed{\text{(a) }4,\qquad \text{(b) }7,\qquad \text{(c) }-3} \]

(a)

We need to find the power to which 3 must be raised to produce 81.

Since:

\[ 3^4=81 \]

we obtain:

\[ \log_3 81=4 \]

(b)

Use the inverse property:

\[ a^{\log_a x}=x \]

Therefore:

\[ 5^{\log_5 7}=7 \]

The exponential and logarithm cancel because they have the same base.

(c)

Rewrite the fraction as a negative power:

\[ \frac18=2^{-3} \]

Therefore:

\[ \log_2(2^{-3})=-3 \]

This question tests the definition of logarithms and the two inverse properties. Make sure you can recognise these relationships without needing a calculator.

Question 2

Solve the equation:

\[ 7^{3x-2}=19 \]

Give your answer correct to three decimal places.

Show your exact logarithmic expression and use the change of base rule to obtain a decimal answer.

Solution 2

The solution is:

\[ \boxed{x\approx1.171} \]

Step 1: Convert the exponential equation into logarithmic form.

\[ 3x-2=\log_7 19 \]

Step 2: Rearrange to make \(x\) the subject.

\[ 3x=2+\log_7 19 \]

Therefore:

\[ x=\frac{2+\log_7 19}{3} \]

Step 3: Apply the change of base rule.

Using natural logarithms:

\[ x=\frac{2+\frac{\ln19}{\ln7}}{3} \]

Step 4: Evaluate using a calculator.

\[ \frac{\ln19}{\ln7}\approx1.513142 \]

Therefore:

\[ x\approx\frac{2+1.513142}{3} \]

\[ x\approx1.171047 \]

Hence:

\[ \boxed{x=1.171} \]

The exact answer is the logarithmic expression, while the decimal answer is an approximation to three decimal places.

Question 3

A student uses a calculator to investigate four expressions.

Classify each exact value as rational or irrational.

For each expression:

(a) State its exact value where possible.

(b) Classify it as rational or irrational.

(c) Justify why \(\log_2 3\) is irrational.

Solution 3

The rational values are \(8\) and \(4\). The irrational values are \(\sqrt2\) and \(\log_2 3\).

The calculator approximations are useful, but exact algebra is needed to justify the classifications.

(a) \(16^{3/4}\)

Rewrite using fractional indices:

\[ 16^{3/4}=(\sqrt[4]{16})^3 \]

Since:

\[ \sqrt[4]{16}=2 \]

we obtain:

\[ 16^{3/4}=2^3=8 \]

This is rational because it is an integer.

(b) \(2^{1/2}\)

We have:

\[ 2^{1/2}=\sqrt2 \]

This is irrational.

Its decimal approximation is:

\[ \sqrt2\approx1.4142 \]

To justify the classification, suppose \(\sqrt2=p/q\), where \(p\) and \(q\) are integers with no common factor.

Squaring gives:

\[ p^2=2q^2 \]

This means \(p\) must be even. Substituting an even value for \(p\) then shows that \(q\) must also be even.

That contradicts the assumption that the fraction is in its simplest form.

Therefore, \(\sqrt2\) is irrational.

(c) \(\log_3 81\)

Since:

\[ 3^4=81 \]

we have:

\[ \log_3 81=4 \]

This is rational.

(d) \(\log_2 3\)

Suppose:

\[ \log_2 3=\frac pq \]

for positive integers \(p\) and \(q\).

Then:

\[ 2^{p/q}=3 \]

Raising both sides to the power \(q\):

\[ 2^p=3^q \]

This is impossible because the two sides have different prime factors.

Therefore:

\[ \boxed{\log_2 3\text{ is irrational}} \]

The important lesson is that a calculator can suggest a value, but a mathematical argument establishes whether it is rational or irrational.

Question 4

Solve the equation:

\[ \log_2(x-1)=\log_4(3x+1) \]

Give an exact answer.

You must:

(a) State the domain restrictions.

(b) Apply the change of base rule.

(c) Solve the resulting equation.

(d) Explain why any algebraic solution outside the original domain must be rejected.

Solution 4

The only valid solution is:

\[ \boxed{x=5} \]

The domain restriction is essential because logarithms cannot have zero or negative arguments.

Step 1: Establish the domain.

For the first logarithm:

\[ x-1>0 \]

Therefore:

\[ x>1 \]

For the second logarithm:

\[ 3x+1>0 \]

Therefore:

\[ x>-\frac13 \]

Combining the restrictions gives:

\[ \boxed{x>1} \]

Step 2: Apply the change of base rule.

We know:

\[ \log_4(3x+1) = \frac{\log_2(3x+1)}{\log_2 4} \]

Since:

\[ \log_2 4=2 \]

the original equation becomes:

\[ \log_2(x-1) = \frac12\log_2(3x+1) \]

Multiply both sides by 2:

\[ 2\log_2(x-1)=\log_2(3x+1) \]

Step 3: Apply the power law.

\[ \log_2((x-1)^2)=\log_2(3x+1) \]

Since logarithms with the same valid base are one-to-one, their arguments must be equal.

Therefore:

\[ (x-1)^2=3x+1 \]

Step 4: Solve the quadratic equation.

Expand:

\[ x^2-2x+1=3x+1 \]

Rearrange:

\[ x^2-5x=0 \]

Factorise:

\[ x(x-5)=0 \]

Hence:

\[ x=0\quad\text{or}\quad x=5 \]

Step 5: Apply the original domain restriction.

We established that:

\[ x>1 \]

Therefore, \(x=0\) is invalid.

The only valid solution is:

\[ \boxed{x=5} \]

Notice the trap: the algebra produces two roots, but only one satisfies the conditions of the original logarithmic equation.

Question 5

Consider the exponential function:

\[ f(x)=\left(\frac1{16}\right)^x \]

and its inverse:

\[ g(x)=\log_{\frac1{16}}x \]

A student makes the following claim:

“Because the two graphs are reflections of each other in the line \(y=x\), every intersection must lie on \(y=x\).”

The claim is incorrect.

(a) Find two distinct intersection points of the graphs that do not lie on \(y=x\). Give exact coordinates and verify both points.

(b) Prove that the two graphs also have an intersection on \(y=x\), with its \(x\)-coordinate lying between 0 and 1.

(c) Explain why the student’s reasoning fails, and identify the important difference between inverse graphs of increasing and decreasing functions.

Solution 5

The two graphs have at least three distinct intersections:

\[ \boxed{ \left(\frac14,\frac12\right),\quad \left(\frac12,\frac14\right),\quad (c,c) } \]

where \(c\) is a number between 0 and 1 satisfying:

\[ \left(\frac1{16}\right)^c=c \]

The key is recognising that reflections can produce pairs of intersections away from the line \(y=x\).

(a) Find two off-diagonal intersections.

Start with:

\[ f(x)=\left(\frac1{16}\right)^x \]

Substitute \(x=\frac14\):

\[ \begin{aligned} f\left(\frac14\right) &=\left(\frac1{16}\right)^{1/4}\\ &=\frac12 \end{aligned} \]

Therefore, the exponential graph contains:

\[ \left(\frac14,\frac12\right) \]

Now consider the logarithmic graph.

\[ \log_{\frac1{16}}\left(\frac14\right)=\frac12 \]

because:

\[ \left(\frac1{16}\right)^{1/2}=\frac14 \]

Therefore, the logarithmic graph also contains:

\[ \left(\frac14,\frac12\right) \]

This is our first intersection.

Now substitute \(x=\frac12\) into the exponential function:

\[ \begin{aligned} f\left(\frac12\right) &=\left(\frac1{16}\right)^{1/2}\\ &=\frac14 \end{aligned} \]

Therefore, the second point is:

\[ \left(\frac12,\frac14\right) \]

Checking the logarithmic function:

\[ \log_{\frac1{16}}\left(\frac12\right)=\frac14 \]

So this point also lies on both graphs.

We have found two distinct intersections:

\[ \boxed{ \left(\frac14,\frac12\right) \text{ and } \left(\frac12,\frac14\right) } \]

Neither point lies on \(y=x\).

(b) Prove that a third intersection exists on \(y=x\).

An intersection on the line \(y=x\) must satisfy:

\[ f(x)=x \]

Therefore:

\[ \left(\frac1{16}\right)^x=x \]

We want to show that this equation has a solution between 0 and 1.

Consider the function:

\[ f(x)=\left(\frac1{16}\right)^x \]

At \(x=0\):

\[ f(0)=1 \]

The exponential curve is above the line \(y=x\), because:

\[ 1>0 \]

At \(x=1\):

\[ f(1)=\frac1{16} \]

The exponential curve is below the line \(y=x\), because:

\[ \frac1{16}<1 \]

The exponential function is continuous, so as it moves from \(x=0\) to \(x=1\), it must cross the line \(y=x\) at least once.

Therefore, there exists some:

\[ 0<c<1 \]

such that:

\[ \boxed{\left(\frac1{16}\right)^c=c} \]

This point lies on the exponential graph.

It also lies on the logarithmic graph because:

\[ \log_{\frac1{16}}c=c \]

Thus, \((c,c)\) is another intersection.

Furthermore, this diagonal intersection is unique because the exponential function is strictly decreasing, while the line \(y=x\) is strictly increasing.

It cannot be either of our earlier points because:

\[ f\left(\frac14\right)=\frac12 \]

and:

\[ f\left(\frac12\right)=\frac14 \]

Neither input equals its output.

Therefore, the three intersections are distinct.

(c) Explain why the student’s reasoning fails.

The mistake is assuming that every intersection of inverse graphs must lie on the line \(y=x\).

Reflection across \(y=x\) means that if a point \((p,q)\) belongs to one graph, the reflected point \((q,p)\) belongs to the other.

It does not mean that the original point must have equal coordinates.

In this example, the two off-diagonal intersections occur as a reflected pair:

\[ \left(\frac14,\frac12\right) \quad\longleftrightarrow\quad \left(\frac12,\frac14\right) \]

This is possible because the exponential function is decreasing.

There is an important difference when the original function is increasing.

For strictly increasing inverse functions, an intersection away from \(y=x\) is impossible. If \(x<y\), an increasing function would require \(f(x)<f(y)\), contradicting the conditions for an off-diagonal intersection of the function and its inverse.

For decreasing functions, that contradiction disappears.

The example demonstrates that two inverse graphs can have off-diagonal intersections as well as an intersection on the line \(y=x\).

This is why the base restriction matters when reasoning about exponential and logarithmic graphs.

10Connecting exponential and logarithmic functions to calculus

The main ideas in this focus area are closely connected.

Exponential functions describe repeated multiplication and provide models for growth and decay. Logarithms reverse exponential operations, allowing us to solve for unknown exponents. The special base \(e\) connects these functions to differentiation because:

\[ \frac{d}{dx}(e^x)=e^x \]

The next important step is applying this relationship to problems involving rates of change.

In particular, understanding \(e^x\) and its inverse \(\ln x\) prepares you for differentiation, continuous growth and decay models, and more advanced applications of calculus in Year 12 Mathematics Advanced.