• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
  6. 6
  7. 7
  8. 8
  9. 9
  10. 10
  11. 11
  12. 12
  13. 13
  14. 14
  15. 15
  16. 16
  17. 17
  18. 18

Type I - Logarithm Bases

Extracts from this document...


Math Portfolio- Logarithm Bases

Consider the following sequences: Find the nth term plus the general statement for each term.

Log2 8, Log4 8, Log8 8, Log16 8, Log32 8, ….

Firstly looking at this sequence the base value is the only number that changes throughout the sequence. The base of the logs in the sequence consists of:

2, 4, 8, 16, 32 …

This value is seems like it is doubling each time. This means that the next two values for the base is going to be:

32 × 2 = 64

64 × 2 = 128

However, when examining the sequence closely the base of the logs cannot be doubling because this means that the nth term is Log2n 8. This is invalid and wrong because if you substitute positive integers for n, the sequence does not appear to be the same. When adding the next two terms to the sequence, the sequence will look as shown:

 Log2 8, Log4 8, Log8 8, Log16 8, Log32 8, Log64 8, Log128 8, …

The nth term for this sequence is Log2n8. This can be simplified and written like this, Log2n23. This proves that the next two base values of the logs are 64 and 128 because 26 = 64 and 27 = 128.

The next sequence is:

Log3 81, Log9 81, Log27 81, Log81 81, …

In this sequence, when examining the base values for each term, the value is tripling; this means that the nth term is Log3n81.

...read more.


. (y = image74.pngimage74.png)image75.png

The graph below represents a graph of Log2n23. (y = Log2n23)image75.png

Analyzing the graphs shown above, you can see that both of the graphs are exactly identical. This fully justifies that both of these equations are exactly the same, and it proves that the expressions for the sequence is equal as well. They are both equal.

The graph below represents a graph of  image55.pngimage55.png. (y = image55.pngimage55.png)


The graph below represents a graph of Log3n34, where y = of Log3n34.image76.png

Analyzing each graph, this is also a justification to the main expression because this follows the concept that the expressions defining the sequences are equal.

The graph below represents a graph of  image68.pngimage68.png. (y = image68.pngimage68.png)image77.png

The graph below represents a graph of Log5n52, where y = of Log5n52.


Looking at these two graphs, again this is obvious that the graphs are identical. Therefore, this proves that the expressions/formulas for the sequence are equal and is valid. Therefore, this means that this is justified and proved.

This method of writing the expression for the log in this form,  image00.pngimage00.png , can then be a way to calculate the actual answer of the Log. The following Logs are answered in this form below:

Log4 64, Log8 64, Log32 64, …

Log4 64 = image78.pngimage78.png

 or Log4 64 = Log2226

= image79.pngimage79.png

=image80.pngimage80.png= image81.pngimage81.png= 3

Log8 64 =image82.pngimage82.png

or Log8 64 = Log2326

= image84.pngimage84.png

= image85.pngimage85.png= image86.pngimage86.png= 2

Log32 64 = image87.pngimage87.png

or Log8 64 = Log2526

= image88.pngimage88.png

= image89.pngimage89.png= image90.pngimage90.png= 1.2

...read more.


1/125 125 = image28.pngimage28.png= image29.pngimage29.png

= This equals,image30.pngimage30.png= image31.pngimage31.png= -1

Log1/625 125 = image33.pngimage33.png= image34.pngimage34.png

= This equals,image35.pngimage35.png= image36.pngimage36.png

Using the other form, to find Log1/625 125, you must:

image37.pngimage37.png=image38.pngimage38.png= image39.pngimage39.png= image40.pngimage40.png

The fourth sequence is:

Log8 512, Log2 512, Log16 512, …

Log8 512 = image41.pngimage41.png= image42.pngimage42.png

This equals,image44.pngimage44.png= image45.pngimage45.png= 3

Log2 512 = image46.pngimage46.png= image47.pngimage47.png

= This equals,image48.pngimage48.png= image49.pngimage49.png= 9

Log16 512 = image50.pngimage50.png= image51.pngimage51.png

= This equals,image52.pngimage52.png= image53.pngimage53.png

Using the other form, to find Log16 512, you must:

image56.pngimage56.png= image57.pngimage57.png = image58.pngimage58.png

Letting Loga X = c and Log b X = d, we can further create a general statement similar to the 2nd way of calculating the logs. (Shown above) (image05.pngimage05.png).

Finding the general statement that expresses Logab X, in terms of c and d is shown in the following steps:

  1. Firstly we know that because of the rules of logarithms, image59.pngimage59.png= image60.pngimage60.png
  2. The second step is to simplify this formula into its simplest form and you should have the complete general statement that expresses Logab X, in terms of c and d.

image60.pngimage60.png= image61.pngimage61.png = image62.pngimage62.png= image63.pngimage63.png= image64.pngimage64.png = image65.pngimage65.png.

Arriving at this general statement relates heavily to the fact about the rules of logarithms. These rules help find the final general statement that expresses the each term and answer for each term.

The Scope and Limitations:

        There are a couple of limitations to this general statement. First is that for the base of all logarithms, you cannot have a negative integer for these values. This means that for a and b the numbers cannot be negative. Furthermore, this shows means that x, a and b must be greater than zero. Another way of expressing this is that the domain of this general statement must be all positive real numbers.

...read more.

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

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Maths essays

  1. Extended Essay- Math

    �]�����5�n�:���{���#Ƨ�}ؤ=A2'�w�}Ç­4��j�%"jW� ��1/2Fp�l"��am�`��9�� �?2>A2�W3f {=�; d����'.�'���6��/�� ���QÛ¶m ��57�H��"+qØ°.+�1/4�E2� zN�=t�P... �<�g�#�"���2-�}�$<A2 :~��d$ t�a�� [�y�d"�< 2b�Y'�9}-?`�)r��# E-F��fE��\"k֬�#�r���n �´ï¿½ï¿½ �s�桴a;O�s$#�p3DV(Ü¿v�e�7�|#3/4_�h'']��T?=='��(c) >l������3`�9V��H(���2M(�v/SN���n�G'']��T�C1/2(tm)�&��}n��&&6���'HF�Ç3|^�VM�9=5n�i[r3/4�E �S ���& î�L C�K''8�`fD2 wHO�ÝvVVVH/�C�d$N%8+ A��cO9`u S�>���$#q�A&#��}yO #F�] �r�HF�T�)�È���0'ݸq��e��B���'#gΜ ;;; +y���-Æ���Z@�g� C�YG���~� jݺ��O?mʺd�'6m� 8pÄΤ��)"G�z��Me(c) ~� #&6'..."�H�'+u-.]Z�LL��֭;|�p�Ç7o�1/4���N�:a���&f q�-r��u���X)sr�C�73a0Å��1�O�>q�D����D(�ܹ�ڵk1/2-��8P�re�21/4w��(tm)�H�w@(tm)(L(tm)2�k�(r)�K-^�~}JJ�YX3/4|y�"�X)��*�^6L1/2���-...'��Å�y�iÓ¦'���'%%(tm)����k7"�"� pr�A(r)-o1/4�FfY""��1/2�##MÆ3w��bÅA��M�->|� K�--[�� #p"��/�Fw�c��4���U����Q(r) �.��Ν;7`���'�-#���6� ��E��h�"3/4 Z��p��V�\%-+W�z��J�ޣ" ��`1��o�3/4�~��ٳ Ò;����"c�P�F���!���"#-0"�."����UWa��}-.]�h�sO-�(r)�3/4fdd��qAea�A!�*K��(��-�{&���k\"V�X �\Ll$ j I&# ����-i�..."�P�0'"WOa5$~����%� r�ر%J$�kz�n1����f���MÎ�`î�"2Q�Hy_AH0��5�'v��� j���6"=�l!�ie��'�<E�q�$#��%� ��4k�,�ڬ�1�a#�Ç�y'��� f�X�J[���w`(tm)��}���K�$�1/4

  2. Arithmetic Sequence Techniques

    Using the conclusions from Question 4, find S3k and S4k. Solution: S0 Sk S2k S3k S4k p q - p q From Question 4, we know that: 2am = ap + aq if m = p + q So: 2(S2k - Sk)

  1. IB Pre-Calculus Logarithm Bases General Information: Logarithms A ...

    + 1 2! + 1 3! + 1 4! + . . . e is an irrational number, whose decimal value is approximately 2.71828182845904. To indicate the natural logarithm of a number, we use the notation "ln." ln x means logex (taken from: http://www.themathpage.com/aPreCalc/logarithms.htm#proof ). After understanding natural logarithms, it makes it easier to understand the expression used when dealing with the sequences mentioned above.

  2. Logarithms. In this investigation, the use of the properties of ...

    log34 / log31 4log3 / 1log3 4/1 2 log981 log32(34) log34 / log32 4log3 / 2log3 4/2 = 2 3 log2781 log33(34) log34 / log33 4log3 / 3log3 4/3 4 log8181 log34(34) log34 / log34 4log3 / 4log3 4/4 = 1 5 log24381 log35(35)

  1. Maths Type I

    The graphs below demonstrate the other parabolas with different values of A and at the bottom is a spreadsheet containing the values for x1, x2, x3, x4, SL and SR and the subsequent values for D. Blue dots represent the intersections between the lines y = x and y = 2x with the parabola.

  2. matrix power

    You then multiply the pairs together and the sum of the products will give a single number which is the first digit of the new matrix. Therefore the matrix equation being solved will look like: Another easier way of solving matrices powers is raise the power of the digits inside

  1. Logrithum bases

    If this pattern is true, I can state that in sequence 4, the nth term in p/q expression is Log m mk / nth. Part III & IV In part three, I will calculate the given logarithms in the form p/q, where p, q?Z.

  2. In this investigation I will be examining logarithms and their bases. The purpose of ...

    To prove this theory I will create two more sequences which follow the same pattern as the sequences above. Then I will evaluate the terms to prove my theory. 1. log6216 log36216 log216216 = 3 = 1.5 or = 1 2.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work