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# Maths Portfolio Type 1

Extracts from this document...

Introduction

Maths Internal Assesment

Logarithm Bases

LOGARITHM BASES

This internal assessment focuses on the logarithms. There are a few rules which govern all the concepts of logarithms:

=c, =b   where a>0, a≠1, b>0

=

• Consider the following sequences. Write down the next two terms of each sequence.

log₂8, log₄8, log₈8, log₁₆8, log₃₂8, log₆₄8, log₁₂₈8

log₃81, log₉81, log₂₇81, log₈₁81, log₂₄₃81, log₇₂₉81

log₅25, log₂₅25, log₁₂₅25, log₆₂₅25, log₆₁₂₅25, log₁₅₆₂₅25

:
, , ,, ,

• Find an expression for the nth term of each sequence. Write down your expression in the form  , where p, q . Justify your answers using technology.

=

Use the = rule.

Then we apply the rule:  = a

We can cross  away on both sides.

What remains is:

We can do this for all the rows.

=

Use the = rule.

Then we apply the rule:  = a

We can cross  away on both sides.

What remains is:

=

Use the =  rule.

Then we apply the rule:  = a

We can cross  away on both sides.

What remains is:

4. Expressed in m, n and k.

Use the =  rule and then we can change the base. We change the base to 10.

Then we apply the rule:  = a

We can cross  away on both sides.

Middle

)

=  = 2

=   ()

=  = 1

=   ()

=  ≈ 0,67

3.  =   ()

=  = -3

=   ()

=  = -1

=   ()

=  = -0,75

4.   =   ()

=  = 3

=   ()

=  = 9

=   ()

=  = 2,25

• Describe how to obtain the third answer in each row from the first two answers. Create two more examples that fit the pattern above.

1.

As we can see, n=2 in the first logarithm and in the second logarithm n=3. If we add these together,  we get n=2+3=5. That means that in the first row, the third answer is obtained by adding the first two n up together. The pattern is therefore that you add up the two n in front of the next logarithm.

The next two examples which would fit in the pattern would therefore be:

, =

2.  =

As we can see, n=1 in the first logarithm and in the second logarithm n=2. There is an arithmetic increase, with the fixed number of 1. The next number in the second row will therefore be n=3. The pattern thus is that there is an arithmetic increase with the fixed number of 1.

The next two examples which would fit the pattern would therefore be:

, =

3.  =

As we can see, n=-1 in the first logarithm and in the second logarithm n=-3.

Conclusion

a>0, b>0, a≠1, b≠1, x>0

We can do a check for this:

Example: a=-2, b=2, x=4

It is impossible to power a function which results in a negative number, in this case a =   -2. With these numbers: , n>0.

The same applies for b, that a>0.

Example 2: a=10, b=1, x=100

N=0 as , which makes it impossible, as you have to divide that number and   = error/not possible.

The sample applies for a, that b≠1.

Example 3: a=4, b=8, c=-8

Same reason as in example 1: it is impossible to power a function which results in a negative number, in this case x (=-8). With these numbers: , k>0.

As a>0 and b>0, the product x should always be greater than 0, therefore x>0.

To sum up again:

a>0, b>0, a≠1, b≠1, x>0

• Explain how you arrived at your general statement.

One law of logarithms state that:

We use the change of base rule

=c then =x                                  =d then =x

==

Take logarithms in base x:

= =

d =

Derived from  we can state that:

+  =

If we change the base again we get the following equation:

We substitute :

The last and following step is to multiply both sides by cd:

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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