(i) $ 6 \times 6 = 6^2$ here $2$ is the index (power) and we read as “six raised to the power two”, or “six squared”

(ii) $ 6 \times 6 \times 6 = 6^3 $ here $3$ is the index (power) and we read as “six raised to the power three”, or “six cubed”

Note: We call the expression that we are raising to the power the base. Here the base is $6$.

Indices appear in formulae for calculating physical quantities such as area and volume.

(i). the area of a square of side length $x$ metres is, $x × x = x^2$ (square metres).

(ii). the area of a circle of radius $r$ centimetres is, $ \pi \; r^2 $ (square centimetres).

(iii). the volume of a cube of side length $y$ metres is, $y × y × y = y^3$ (cubic metres).

(iii). the volume of a sphere of radius $r$ centimetres is, $ {\Large\frac{4}{3}} \; \pi \; r^3$ (cubic centimetres).

Scientific notation provides us with an easy way for writing very large, or very small, numbers. In scientific notation we have:

(i). $ 3600000 = 3.6 \times 10^6 $

(ii). $ 0.0051 = 5.1 \times 10^{-3}$

(iii). $ 0.000078 = 7.8 \times 10^{-5}$

A computer algorithm that loops as follows:

for $i = 1$ to $n$

for $j = 1$ to $n$

<perform action>

next $j$

next $i$

goes around the loop $n \times n$ times = $n^2$ times.

Indices therefore feature in a wide range of day-to-day calculations and there are associated rules that apply when performing calculations involving indices.

(i).    $ a^3 \times a^2 = a^{3+2} = a^5$

(ii).  $ 4^9 \times 4^6 = 4^{9+6} = 4^{15} $

(iii).   $ 2^6 \times 2^{11} = 2^{6+11} = 2^{17} $

in general   $ \class{outbox}{a^m \times a^n = a^{m+n}} $

(i).   $ {\Large\frac{a^6}{a^4}} = a^{6-4} = a^2 $

(ii).   $ {\Large\frac{5^{16}}{5^7}} = 5^{16-7} = 5^9 $

(iii).   $ {\Large\frac{7^9}{7^3}} = 7^{9-3} = 7^6 $

in general,    $ \class{outbox}{{\Large\frac{a^m}{a^n}} = a^{m-n}}$

(i).   $ (a^3)^2 = a^{3 \times 2} = a^6 $

(ii).   $ (3^8)^7 = 3^{8 \times 7} = 3^{56} $

(iii).   $ (4^5)^6 = 4^{5 \times 6} = 4^{30} $

   $ \class{outbox}{(a^m)^n = a^{m \times n} = a^{mn}}$

This law can actually be derived from Law1. For example from (i) above,

$$ (a^3)^2 = a^3 \times a^3 = a^{3+3} = a^6. $$

Hence, we are able to simplify expressions like the square of a cube.

$ {\Large\frac{a^2}{a^2}} = a^{2-2} = a^0$ by law 2. $ (a \neq 0) $

However, $ {\Large\frac{a^2}{a^2}} = 1 $ as any number (except $0$) divided by itself equals $1$.

Hence, we must have that

$ \class{outbox}{a^0 = 1, \; a \neq 0 }$

$ (2 \times 4)^3 = (2 \times 4) \times (2 \times 4) \times (2 \times 4) $

$ = (2 \times 2 \times 2) \times (4 \times 4 \times 4) $

$ =\;$$2^3\times4^3$

in general    $ \class{outbox}{(ab)^n = a^nb^n }$

$$ \Big(\frac{2}{3}\Big)^4 = \frac{2 \times 2 \times 2 \times 2}{3 \times 3 \times 3 \times 3} = \frac{2^4}{3^4} $$

In general

$$ \class{outbox}{\Big(\frac{a}{b}\Big)^n = \frac{a^n}{b^n}} $$

Hence, we apply the power to both expressions.

Use the laws of indices to simplify each of the following:

(i). $ 2^2 \times 2^5 = 2^{2+5} = 2^7 = 128 $

(ii). $ {\Large\frac{2^4}{2^3}} \times 2^{4-3} = 2^1 = 2$

(iii). $ (4^2)^2 = 4^{2 \times 2} = 4^4 = 256$

(iv). $ \Big({\Large\frac{2}{3}}\Big)^3 = {\Large\frac{2^3}{3^3}} = {\Large\frac{8}{27}}$

(v). $ (4 \times 5)^3 = 4^3 \times 5^3 = 64 \times 125 = 8000 $

(vi). $ {\Large\frac{6^3 \times 6^4}{6^7}} = {\Large\frac{6^{3+4}}{6^7}} = {\Large\frac{6^7}{6^7}} = 6^{7-7} = 6^0 = 1 $

(vii). $ (2x)^3 = 2^3x^3 $

(viii). $ \Big({\Large\frac{3x}{4y}}\Big)^2 = {\Large\frac{3^2x^2}{4^2y^2}} = {\Large\frac{9x^2}{16y^2}} $

(ix). $ (3x^3y^4)(4x^2y^5) = 3x^3y^44x^2y^5 $

$ = 12x^{3+2}y^{4+5} $

$= 12x^5y^9$

(x). $ (2a^2b^3c)^4 \;\; = (2)^4(a^2)^4(b^3)^4(c^1)^4 $

$ =2^4a^{2 \times 4}b^{3 \times 4}c^4 $

$ = 16a^8b^{12}c^4$

(xi). $ {\Large\frac{((2x)^2)^3(2)}{2^7x^6}} = {\Large\frac{(2^2x^2)^3(2)}{2^7x^6}} $

$ = {\Large\frac{(2^6x^6)(2)}{2^7x^6}} $

$ = {\Large\frac{2^7x^6}{2^7x^6}} $

$ = $$\class{double}{1} $

We now turn our attention to look at cases where the exponents are negative and/or fractional.

$$ \frac{a^2}{a^5} = \frac{a \times a}{a \times a \times a \times a \times a} = \frac{1}{a \times a \times a} = \frac{1}{a^3} \;\;\; (a \neq 0) $$

However, from Law 2 we have that $ {\Large\frac{a^2}{a^5}} = a^{2-5} = a^{-3} $

Hence, we must have that $ {\Large\frac{1}{a^3}} = a^{-3}$

in general   $ \class{outbox}{{\Large\frac{1}{a^n}} = a^{-n} (a \neq 0).} $

or equivalently  $ \class{outbox}{{\Large\frac{1}{a^{-n}}} = a^n (a \neq 0).} $

This law enables us to move expressions from the denominator to the numerator.

(i). From Law 1 we can write $ a^{1/2} \times a^{1/2} = a^1.$

We know that $ \sqrt{a} \times \sqrt{a} = a^1$

Hence, we conclude that $ a^{1/2} = \sqrt{a} $

The symbol $\sqrt{a} $ represents the square root of $a$, i.e. the number that when multiplied by itself returns the original number, $a$.

Note however that every number has a positive and a negative square root. For example,

$ \sqrt{4} \times \sqrt{4} = 2 \times 2 = 4 $ and $ \sqrt{4} \times \sqrt{4} = (-2) \times (-2) = 4$.

In our work, unless otherwise stated, we will work with the positive square root.

(ii). From Law 1 we can write $ a^{1/3} \times a^{1/3} \times a^{1/3} = a^1 $.

We know that $ \sqrt[3]{a} \times \sqrt[3]{a} \times \sqrt[3]{a} = a^1 $.

Hence, we conclude that $ a^{1/3} = \sqrt[3]{a} $, i.e. the cube root of $a$.

In general,    $ \class{outbox}{a^{1/n} = \sqrt[n]{a}}$

The number in the denominator of the index specifies the root we are calculating.

(i). Now consider $ a^{1/3} \times a^{1/3} = (a^{1/3})^2 = a^{2/3}$.

Also, $ \sqrt[3]{a} \times \sqrt[3]{a} = (\sqrt[3]{a})^2$.

Hence, we conslude that $ a^{2/3} = (\sqrt[3]{a})^2 $

The expression, $a^{2/3}$ means the square of the cube root of $a$.

Note that in the exponent the number in the denominator (3) is the root and the number in the numerator (2) is the power.

in general    $ \class{outbox}{a^{m/n} = (a^{1/n})^m = (\sqrt[n]{a})^m }$   or    $ \class{outbox}{ a^{m/n} = (a^m)^{1/n} = \sqrt[n]{a^m}}$.

(i). $ {\Large\frac{4x^2y^{-6}}{3x^{-5}y^6}} = {\Large\frac{4}{3}}(x^2y^{-6})(x^5y^{-6}) $

$ ={\Large\frac{4}{3}} x^{2+5}y^{-6-6} $

$ ={\Large\frac{4}{3}}x^7y^{-12} $

$ ={\Large\frac{4x^7}{3y^{12}}}$

(ii). $ {\Large\frac{2p^4q^{-3}}{7p^{-2}q^{-6}}} = {\Large\frac{2}{7}}(p^4q^{-3})(p^2q^6) $

$ = {\Large\frac{2}{7}}p^{4+2}q^{-3+6} $

$ = {\Large\frac{2}{7}}p^6q^3$

(iii). $ \Big({\Large\frac{5x^6y^2}{x^2y^{-5}}} \Big)^4 = {\Large\frac{5^4x^{6 \times 4}}y^{2 \times 4}}{x^{2 \times 4}y^{-5 \times 4}} $

$ = {\Large\frac{5^4x^{24}y^8}{x^8y^{-20}}} $

$ = 5^4x^{24}y^8x^{-8}y^{20} $

$ = 625x^{24-8}y^{8+20} $

$ = 625x^{16}y^{28}$

(iv). $ r^4s^{-7}(6r^2s)^{-3} = r^4s^{-7}6^{-3}r^{-6}s^{-3} $

$ = r^{4-6}s^{-7-3}6^{-3} $

$ = r^{-2}s^{-10}6^{-3} $

$ = {\Large\frac{1}{r^2s^{10}6^{3}}} $

$ = {\Large\frac{1}{216r^2s^{10}}} $

(v). $ {\Large\frac{p^{1/4}q^{4/3}r^2}{p^{3/4}q^{1/3}r^{1/2}}} = p^{1/4}p^{-3/4}q^{4/3}q^{-1/3}r^2r^{-1/2} $

$ = p^{-1/2}q^1r^{3/2} $

$= {\Large\frac{qr^{3/2}}{p^{1/2}}}$

Evaluate each of the following expressing your answer as an exact numerical value.

(i). $ 8^{2/3} = (8^2)^{1/3} = \sqrt[3]{8^2} = \sqrt[3]{64} = \class{double}{4} $

(iii). $ 27^{2/3} = (3^3)^{2/3} = 3^{3 \times 2/3} = 3^2 = \class{double}{9} $

Examples (i) - (iii) produced exact solutions in the form of whole numbers or fractions. However, a number such as $ \sqrt{7} = 7^{1/2} $ does not produce an exact solution and evaluates to give an infinite decimal which to four decimal places is $2.6456$. We now take a brief look at numbers of this form.

Solve the following equations for $x$.

(i). $ 9^{x-1} = 3^{3x+1}$

Both sides of the equation must be expressed in terms of the same base number.

Then compare the exponents and solve for $x$. We have

$9^{x-1} = 3^{3x+1} $

$(3^2)^{x-1} = 3^{3x+1} $    (base number on both sides must be 3)

$3^{2x-2} = 3^{3x+1}$    (both sides now have 3 as the base number)

$2x - 2 = 3x + 1$    (equate exponents)

$x = -3$   (solve the $x$)

(ii). $6^{2x-1} = 1 $

$ 6^{2x-1} = 6^0 $    (base number on both sides is now 6)

$ 2x - 1 = 0 $    (equate exponents)

$x = 1 ∕ 2$  (solve the $x$)

(iii). $ {\Large\frac{3^x}{81}} = 81 $

$ 3^x = 81^2 $

$ 3^x = (3^4)^2$

$ 3^x = 3^8 $

$x = 8$

(iv). $2^{x^2-7} = 4^{x + 4} $

$2^{x^2-7} = (2^2)^{x+4} $

$2^{x^2-7} = 2^{2x+8} $

$ x^2 - 7 = 2x + 8 $

$ x^2 - 2x - 15 = 0 $

$ (x - 5)(x + 3) = 0 $

$x = 5$ or $x = -3$

(v). $ \Big({\large\frac{1}{3}}\Big)^{2x+7} = 243 $

$ (3^{-1})^{2x+7} = 3^5 $

$ 3^{-2x-7} = 3^5$

$ -2x - 7 = 5 $

$x = - 6$

(vi). $ \Big(\frac{1}{16}\Big)^{x-3} - 23 = 1001 $

$ \Big({\Large\frac{1}{4^2}}\Big)^{x-3} = 1024$

$(4^{-2})^{x-3} = 4^5$

$4^{-2x+6} = 4^5$

$-2x + 6 = 5$

$-2x = -1$

$x = 1 ∕2$

The money in an account grows by 4% per annum. Assume the account starts with £2000.

(i). Derive a general expression for the amount in the account in $n$ years.

(ii). How much will be in the account in 10 years?

Solution

(i). We can derive an expression for the amount in the account after a given number of years by generalising from a few cases.

Let $x_n$ be the amount in the account at the beginning of year $n$.

Choose the initial year to be year $0$ so that $x_0 = 2000$.

The amount in the account at the beginning of year 1 is,

$x_1 = 2000 + 0.04 \times 2000 = (1 + 0.04) \times 2000 = 2000(1.04)$.

This is the initial amount in the account plus the accrued interest.

Now consider the amount at the beginning of year 2.

$x_2 = x_1 + 0.04x_1 = 1.04x_1$

We already have an expression for $x_1$, and so

$x_2 = (1.04)(1.04)2000 = 2000(1.04)^2$

This generalises to $x_n = 2000(1.04)^n$.

We have derived an indicial formula for the amount in the account at the start of year $n$:

$x_n = 2000(1.04)^n$.

(ii). To find the amount in the account after 10 years we set $n = 10$ in the formula derived in part (i), i.e.

$x_{10} = 2000(1.04)^{10} = $$\class{double}{£}$$\class{double}{2}$$\class{double}{9}$$\class{double}{6}$$\class{double}{0}$$\class{double}{.}$$\class{double}{4}$$\class{double}{9}$.

The above type of development is common in indicial modelling – you should make sure that you are comfortable with it.

A newly discovered oilfield is believed to contain 10 billion barrels of crude oil. It will have oil removed from it at the rate of 9% per annum. Derive an expression for the amount of oil remaining in the field after $n$ years and use it to calculate how much will be in the field after 8 years. How much oil will be removed in the ninth year?

Solution

Denote the amount of oil (barrels) in the field at the end of year $n$ by $A_n$. For convenience we work in units of billions.

$A_0 = 10 $

$A_1 = (0.91)10$   (we have used $A_1 = A_0 - 0.09A_0 = (0.91)A_0$)

$A_2 = (0.91)^210$    (we have used $A_2 = A_1 - 0.09A_1 = 0.91A_1$
$= (0.91)(0.91)A_0 = (0.91)^2A_0)$

In general we obtain an expression for the amount of oil remaining in the field after $n$ years as,

$A_n = (0.91)^nA_0$.

After 8 years the quantity $A_8 = (0.91)^8(10) = 4.70$ billion barrels. This is the amount of oil remaining in the field after 8 years.

After 9 years, $A_9 = (0.91)^9(10) = 4.28$ billion barrels remain in the field.

Hence, the quantity extracted during year 9 is the difference between these two values, i.e. $4.70 - 4.28 = 0.42$ billion barrels. (An alternative for this problem would be to take 9% of $A_8$).

Plot the functions, $ y = 2^x $ and $ y = 2^{-x} = {\Large\frac{1}{2^x}} $ on the same graph.

Solution

Using a calculator for selected $x$ values we create the table

Now plot the graphs but restrict the $x$-interval to between $– 2$ and $2$ for clarity.

Note the following:

Both functions are positive for all values of $x$.

Both graphs pass through the point $(0, 1)$.

The graphs are mirror images of each other in the $y$-axis.

The graph of $y = 2^x$ increases without bound with increasing $x$, i.e. it exhibits exponential growth.

The graph of $y = 2^{-x}$ decreases rapidly with increasing $x$, i.e. it shows exponential decay. Note that the graph never reaches the $x$-axis.

(i). The figure below shows the graphs of several exponential functions, $y = a^x, \; (a > 1)$.

Properties to note from the graph are:

As $a > 1$ the functions all display exponential growth.

The functions are positive for all values of $x$.

In all cases as $x$ increases, $y$ grows very fast and keeps on growing.

All graphs pass through the point $(0, 1)$.

One exponential function that occurs frequently when modelling real world phenomena uses the constant $e = 2.7182818...$ as its base. It occurs so often that it is often referred to as THE exponential function. The graph of $y = e^x$ is shown, as expected, sandwiched between the graphs of $y = 2^x$ and $y = 3^x$ as $2 < e < 3$.

(ii). The figure below shows the graphs of several exponential functions, $y = a^{-x}, (a > 1)$.

Properties to note from the graph are:

As $a > 1$ the functions all display exponential decay.

In all cases as $x$ increases, $y$ decreases rapidly, and keeps on decreasing, but never actually reaches $y = 0$, i.e. the $x$-axis.

The functions are positive for all values of $x$.

All graphs pass through the point $(0, 1)$.

Suppose that the number of bacteria in a certain culture doubles every hour. Assuming that the culture has 1000 bacteria to start with:

(i). determine a formula that gives the number of bacteria, $N(t)$, in the culture after $t$ hours.

(ii). determine the number of bacteria in the culture after 4 hours.

Solution

Here we assume that the bacteria have enough space and a food source to assist continued growth.

(i). Let $N(t)$ denote the number of bacteria in the culture after $t$ hours

After $t = 0\; \text{hours}: N(0) = 1000$

After $t = 1\; \text{hour}: N(1) = (1000)(2) = 2000$

After $t = 2\; \text{hours}: N(2) = [(1000)(2)](2) = (1000)(2^2) = 4000$

After $t = 3\; \text{hours}: N(3) = [(1000)(2)(2)](2) = (1000)(2^3) = 8000$

Examining the structure we have:

$N(0) = 1000 = (1000)(2^0)$

$N(1) = 2000 = (1000)(2^1)$

$N(2) = 4000 = 1000 (2^2)$

$N(3) = 8000 = 1000 (2^3)$

Hence, $N(t) = 1000(2^t)$ giving the general formula for the number of bacteria in the culture after $t$ hours.

(ii). After 4 hours the number of bacteria in the culture is: $N(4) = 1000(2^4) = 16000$.

On a particular island there are currently 1000 rodents. Various studies have shown that the size of this population has been decreasing at an average rate of 5% per annum for the past decade.

(i). determine a formula for $N(t)$, the number of rodents after $t$ years.

(ii). determine how many rodents will be on the island in 10 years time.

Solution

Here we assume that the rate of decline of 5% will continue.

(i). If 5% of the population is lost after one year then 95% of the population is left.

Let $N(t)$ denote the number of rodents on the island after $t$ years.

After $t = 0\; \text{years}: N(0) = 1000$

After $t = 1\; \text{year}: N(1) = (1000)(1 - 0.05)$

$= (1000)(0.95)$

After $t = 2\; \text{years}: N(2) = N(1)(1 - 0.05)$

$= N(1)(0.95)$

$= (1000)(0.95)(0.95)$

$= (1000)(0.95)^2$

After $t = 3\; \text{years}: N(3) = N(2)(1 - 0.05)$

$= N(2)(0.95)$

$= (1000)(0.95)^2(0.95)$

$= (1000)(0.95)^3$

Continuing in this way the number of rodents in the population after $t$ years is given by

$N(t) = (1000)(0.95)^t$.

(ii). The number of rodents after 10 years is given by:

$N(10) = (1000)(0.95)^{10}$
$= 599$ to the nearest whole number.

Assuming that the annual average decline remains at 5%, there will be 599 rodents on the island in 10 years.

A bacterial culture follows a growth model given by

$P(t) = P(0)e^{0.2554t}$

where $P(0)$ is the initial population of 3000 and $P(t)$ is the size of the population at time $t$ hours.

(i). Determine the population size at time $t = 4$ hours.

(ii). If the culture medium can only support a population of 50000 determine the time, $T$ say, when the population reaches this value.

Solution

(i). Here $P(0) = 3000$ and $t = 4$ so the formula becomes

$P(4) = 3000e^{0.2554\times4}$

$= 8333$ bacteria to the nearest whole number.

Hence, after 4 hours the population size has grown to approximately $8333$ bacteria.

(ii). Setting $t = T$ we require the value of $T$ for which $P(T) = 50000$.

Hence, $50000 = 3000e^{0.2554T}$ which simplifies to give

$e^{0.2554T} = {\Large\frac{50}{3}}$.

Here we need knowledge of logarithms to solve for $T$ and this topic will be presented in the next unit.