- Level: International Baccalaureate
- Subject: Maths
- Word count: 1056
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.
Found what you're looking for?
- Start learning 29% faster today
- 150,000+ documents available
- Just £6.99 a month