There is a better way to change bases.

We can rewrite {jatex options:inline}log_a x^y{/jatex} as {jatex options:inline}y log_a x{/jatex}.

There is a similar rule for changing the base.

{jatex options:inline}log_{a^b} x = \frac{1}{b} log_a x{/jatex}

Suppose we want to change the base from an arbitrary base {jatex options:inline}u{/jatex} to some other arbitrary base {jatex options:inline}v{/jatex}.

We can use {jatex options:inline}u= v^{\frac{ln u}{ln v}}{/jatex} where {jatex options:inline}ln u{/jatex} is the natural logaithm of {jatex options:inline}u{/jatex} that is {jatex options:inline}ln u = log_e u{/jatex} where {jatex options:inline}e=2.71828...{/jatex}

Then {jatex options:inline}ln_u x = ln_{v^{\frac{ln u}{ln v}}} x = 1/ \frac{ln u}{ln v} ln_v x = \frac{ln v}{ln u} ln_v x{/jatex}

Use of the base {jatex options:inline}e{/jatex} here is optional. Any base will do, but natural logs are commonl;y used.

Example: Change the base pf a log vrom 7 to 12.

{jatex options:inline}log_7 x = log_{12^{ln 7}{ln 12}} x = \frac{ln 12}{ln 7} ln_{12} x {/jatex}]]>

Example: The first three terms of a geometric sequence areandFindthe first term and the common ratio.

The common ratio is equal to the second term divided by the first and also equal to the third term divided by the second, hence

Cross multiplication gives

Expanding both sides and simplifying gives

Hence

soor

Ifthe first three terms are 1,-2, 4. The first term is 1 and the common ratio is -2

Ifthe first three terms are 5, 10, 20. The first term is 5 and the common ratio is 2.

Example: The first first, second and fourth terms of a geometric sequence are the first, second and third terms of an arithmetic sequence. Find the common ratio of the geometric sequence.

The first four terms of the geometric sequence may be writtenso the first second and fourth terms are

Since these are the first, second and third terms of an arithmetic sequence,

Movingto the right hand side,becomes a common factor so we can factorise with

Henceoris a root of

Ifthen the terms of the sequence are all the same. The common difference is then 0.

Ifthen the terms of the sequence will alternate in sign so there cannot be a single number added to each term to give the next term, so

]]>From the diagram, {jatex options:inline}\theta = tan^{-1} ( \frac{4}{3} ){/jatex} and {jatex options:inline}\theta - \beta = sin^{-1} ( \frac{2.5}{5} ) {/jatex} then {jatex options:inline}\beta = \theta - (\theta - \beta ) = tan^{-1} ( \frac{4}{3} )- sin^{-1} ( \frac{2.5}{5} )=23.13^o {/jatex}]]>

From the Cosine Rule, for the triangle CAB,and for the triangle OAB,

Setting these equal gives

Now use the identity cos 2x = 1-2 sin^2 x rearranged asto give

Dividing by 4 and square rooting gives

]]>Draw the chord AB then A is the sum ofandin the diagram below.

The formula for the area of a segment subtending an anglein a circle of radius x is

Henceandso

]]>The quadrilateral is a kite since sides with the same length pair connect.

The area of a kite is

The diagonal AB is given using the cosine rule.

hence

So

]]>2, 6, 10, 14, 18

has a constant common difference term 4. We add 4 to each term to get the next term. We can write down the rule:

Arithmetic sequences can be defined iteratively, so that each term is calculated from the last term, or in closed form, such that we have a formula for the n ^{th } term.

For the above sequence the closed form would beIn general the closed form for the nth term can be found from the expressionwhere is the first term andis the common difference. There is also a formula for the sum of the first n terms:The fomulae may be used in the following ways:

A sequence starts 5, 9, 13, 17, 21. Find the sum of all the terms between 100 and 200.

the first termis 5 and the common differenceis 4. We need to find how many terms are less than 100 and how many are less than 200.

terms are less than or equal to 100.

terms are less than or equal to 200.

The difference ofandwill give the sum of all the terms between 100 and 200, so the answer is 4949-1224=3725

Example: for the sequence 5, 12, 19, 26, 33 find the sum of all the terms less than 200.

The first termis 5 and the common differenceis 7.

]]>Bearings involve using trigonometry, generally the cosine or sine rules:

Cosine Rule:

Sine Rule:

For the above diagram, find

a)The distance BC

b)The bearing of A from B and the bearing of A from C.

a)Label the triangle as above, with sides labelled by little letters opposite angles labelled by big letters.

so

b)

We draw a right angled triangle between A and B. Construct a right angled triangle at A and a horizontal line starting from A to the right.The internal angles of the right angled triangle are 70 and 20 degrees. Hence the bearing of A from B is 360-20=340 degrees.

To find the bearing of A from C, use the Sine Rule to find C.

Construct the right angled triangle as shown, with the angles at C then the bearing is 56+32.75=88.75 degrees.

]]>It is quite tricky to calculate the month repayment. The method is illustrated in this example.

A loan of £500 is taken out at 11% annual interest to be repaid at a fixed monthly rate. Interest is to be charged from the date of the loan and the first repayment is to be made a month after the loan is taken out. Find the monthly repayment.

Call the monthly repayment x. Since the annual rate of interest is 11%, the monthly rate of interest is the solution to

After a month the amount owing in £ isand after the first monthly payment the amount owing is

After two months the amount owing in £ isand after the second monthly payment the amount owing is

After three months the amount owing in £ isand after the second monthly payment the amount owing is

We will carry on in the same way, until after two years, twenty four monthly repayments, the whole loan is paid off. After the last monthly payment there is no money left owing so

The expression on the right is a geometric series with first termand common ratioso the right hand side equalsand

Hence

Substituting the value ofinto this expression gives a monthly repayment of £23.18 to the nearest penny. The total amount repaid is £556.41 to the nearest penny.

]]>We are only interested in the coefficient of {jatex options:inline}x^4{/jatex} so ignore any powers of {jatex options:inline}x{/jatex} higher than {jatex options:inline}x^4{/jatex}.

{jatex options:inline}\begin{equation} \begin{aligned} (3-x)^6(3+x)^4 &= (9-6x_x^2)(9-x^2)^4 \\ &= (9-6x+x^2)({}^4C_09^4(-x^2)^0+{}^4C_1 9^3(-x^2)^1+{}^4C_2 9^2(-x^2)^2+ higher \; powers \; of \; x) \\ &= (9-6x+x^2)(6561-2916x^2+486x^4 + higher \; powers \; of \; x ) \end{aligned} \end{equation}{/jatex}

The only contributions to the coefficient of {jatex options:inline}x^4{/jatex} are from {jatex options:inline}x^2 \times -2916 x^2{/jatex} and {jatex options:inline}9 \times 486x^4=4374x^4{/jatex}. The coefficient is {jatex options:inline}-2916+4374=1458{/jatex}.]]>