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Maths Portfolio Type 1

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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:

image00.pngimage00.png=c, image49.pngimage49.png=b   where a>0, a≠1, b>0

image88.pngimage88.png= image94.pngimage94.png

  • 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

:
image173.pngimage173.png, image01.pngimage01.png, image14.pngimage14.png,image23.pngimage23.png, image32.pngimage32.png, image41.pngimage41.png


  • Find an expression for the nth term of each sequence. Write down your expression in the form image12.pngimage12.png , where p, q image13.pngimage13.png. Justify your answers using technology.

image66.pngimage66.png= image77.pngimage77.png

Use the image88.pngimage88.png= image94.pngimage94.pngrule.

image103.png

Then we apply the rule: image108.pngimage108.png = a image117.pngimage117.png

image127.png

We can cross image131.pngimage131.png away on both sides.

What remains is: image142.pngimage142.png

We can do this for all the rows.

image154.pngimage154.png= image157.pngimage157.png

Use the image88.pngimage88.png= image94.pngimage94.pngrule.

image161.png

Then we apply the rule: image108.pngimage108.png = a image162.pngimage162.png

image163.png

We can cross image164.pngimage164.png away on both sides.

What remains is: image165.pngimage165.png

image166.pngimage166.png= image167.pngimage167.png

Use the image88.pngimage88.png= image94.pngimage94.png rule.

image168.png

Then we apply the rule: image108.pngimage108.png = a image162.pngimage162.png

image169.png

We can cross image170.pngimage170.png away on both sides.

What remains is:  image171.pngimage171.png

4. Expressed in m, n and k.

image172.png

Use the image88.pngimage88.png= image94.pngimage94.png rule and then we can change the base. We change the base to 10.

image174.png

Then we apply the rule: image108.pngimage108.png = a image162.pngimage162.png

image175.png

We can cross image176.pngimage176.png away on both sides.

...read more.

Middle

)

image31.pngimage31.png = image33.pngimage33.png = 2

image34.pngimage34.png = image35.pngimage35.png  (image03.pngimage03.png)

image36.pngimage36.png = image37.pngimage37.png = 1

image38.pngimage38.png = image39.pngimage39.png  (image03.pngimage03.png)

image40.pngimage40.png = image42.pngimage42.png ≈ 0,67

3. image43.pngimage43.png = image44.pngimage44.png  (image03.pngimage03.png)

image10.pngimage10.png = image45.pngimage45.png = -3

image46.pngimage46.png = image47.pngimage47.png  (image03.pngimage03.png)

image48.pngimage48.png = image50.pngimage50.png = -1

image51.pngimage51.png = image52.pngimage52.png  (image03.pngimage03.png)

image53.pngimage53.png = image54.pngimage54.png = -0,75

4.  image55.pngimage55.png = image56.pngimage56.png  (image03.pngimage03.png)

image57.pngimage57.png = image58.pngimage58.png = 3

image59.pngimage59.png = image60.pngimage60.png  (image03.pngimage03.png)

image61.pngimage61.png = image62.pngimage62.png = 9

image63.pngimage63.png = image64.pngimage64.png  (image03.pngimage03.png)

image65.pngimage65.png = image67.pngimage67.png = 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.  image68.pngimage68.png

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:

image69.pngimage69.png, image70.pngimage70.png= image71.pngimage71.png

2. image72.pngimage72.png = image73.pngimage73.png

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:

image74.pngimage74.png, image75.pngimage75.png= image76.pngimage76.png

3. image78.pngimage78.png = image79.pngimage79.png

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

...read more.

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

image146.png

image147.png

It is impossible to power a function which results in a negative number, in this case a =   -2. With these numbers: image148.pngimage148.png, n>0.

The same applies for b, that a>0.

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

image149.png

image150.png

N=0 as image151.pngimage151.png, which makes it impossible, as you have to divide that number and  image152.pngimage152.png = error/not possible.

The sample applies for a, that b≠1.

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

image153.png

image155.png

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: image156.pngimage156.png, 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:

image93.png

We use the change of base rule

image89.pngimage89.png=c then image49.pngimage49.png=x                                  image90.pngimage90.png=d then image95.pngimage95.png=x

image96.pngimage96.png=image97.pngimage97.pngimage98.pngimage98.png=image97.pngimage97.png

Take logarithms in base x:

image99.pngimage99.png=image100.pngimage100.pngimage101.pngimage101.png = image100.pngimage100.png

image158.pngimage158.pngimage159.pngimage159.pngimage104.pngimage104.pngimage105.pngimage105.png d =  image106.pngimage106.pngimage107.pngimage107.png

Derived from image93.pngimage93.png we can state that:

image104.pngimage104.png + image107.pngimage107.png = image109.pngimage109.png

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

image160.png

We substitute image111.pngimage111.png:

image112.png

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

image113.png

...read more.

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

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