• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

IB SL Math Portfolio- Logarithm Bases

Extracts from this document...

Introduction

IB Math SL Portfolio Type 1


General Introduction:

In its simplest terms, a logarithm is an exponent. It is an exponent needed to produce a given number from a specific base. It is written in the form loggh=j, which denotes gh=j (g to the power of h equals j).

LOGARITHM BASES

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

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

        log381, log981, log2781, log8181, … log24381, log72981, ...

        log525, log2525, log12525, log62525, … log3,12525, log15,62525, ...

        :

        :

        :

        logmmk, logm^2mk, logm^3mk, logm^4mk, … logm^5mk, logm^6mk, …

Find an expression for the nth term of each sequence. Write your expressions in the form p/q, where p and q are integers. Justify your answers using technology.

The first sequence can be otherwise written:

log2^123, log2^223 , log2^323, log2^423, log2^523, ...

Here, I noticed a pattern: the base of each logarithm is 2n. Using this knowledge and the concepts of the change of base rule and verifying my theories with a GDC calculator, I developed the expression shown here by evaluating each term.

(Each term to be evaluated can be checked by using a GDC calculator. For example, log28

...read more.

Middle

25, log5^325, log5^425, …

log5^152= 2/1

log5^252= 2/2

log5^352= 2/3

log5^452 = 2/4

:

:

log5^n52 = 2/n

The nth term of the third sequence can expressed as 2/n.

The same rules apply for the fourth and final sequence.

logm^1mk, logm^2mk, logm^3mk, logm^4mk, …

logm^1mk= k/1

logm^2 mk= k/2

logm^3 mk= k/3

logm^4 mk= k/4

:

:

logm^n mk= k/n

The nth term of the fourth sequence can expressed as k/n.


Now, calculate the following, giving your answers in the form p/q, where p and q are both integers.

        log464,        log864,           log3264                3/1,        2/1,        6/5

        log749,          log4949,          log34349                2/1,        2/2,        2/3

        log1/5125,         log1/125125,         log1/625125                3/-1,        3/-3,        3/-4

        log8512,         log2512,         log16512                3/1,        9/1,        9/4

Describe how to obtain the third answer in each row from the first two answers.

First, I took the lowest prime base possible of each.

log2^264,        log2^364,           log2^564

        log7^149,          log7^249,          log7^349

        log(1/5)^1125,         log(1/5)^3125,         log(1/5)^4125

        log2^3512,         log2^1512,         log2^4512

In doing this, I

...read more.

Conclusion

logax=clogbx=d

log864=4                        log464=2

c=4                                d=3

logabx= cd/(c+d)

        log3264= 6/5

or

        326/5=64

Discuss the scope and/or limitations of a, b and x.

The variables a and b must be integers greater than zero and not equal to one. If a or b were equal to one, then c and d would be rendered undefined.

(For example, if a=1 and x=3, the given statement logax=cwould be undefined for c

because there is no exponent that can be raised to make one equal three. One to any power always equals one.)

This is the same for b. If it were equal to one, d would be undefined.

They cannot be negative, as this does not work with the equations.

The variable c cannot equal negative d, as this would yield a zero in the denominator of the equation, which would make it undefined.

Explain how you arrived at your general statement.

I explained it step by step above, but generally I just used my knowledge of logarithm and exponent rules to mold and simplify the information I was given. For example, I knew that logax=c was the same as writing x= ac, a point I explained in the brief introduction,which helped me not only with this portion but the portfolio in its entirety.


...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

    "W�(c)eV'�v�k���>[\L�ZJ�A!Q@it�D���@"6d�B���MO�7M�o�8%g�...3eH/ �s1/4����M� x�);-�˱E�@�S",Z(c)^Z�?);ȡ�MZD��^T�(c)^��F��A7m��2�R"~D�ß���:�Ѭ��d=2O��ۤ7�(c)'S��'|���OD~z ��o���NGy��ڨ��["8�?|koqc�"3/4�p����U(r)�)"911/4DN�*�mC=� ���� 8���G��� �1/2k(c)M ������G�X�w��I��U��;��� ��/�"�1/4� ��x��{�Nj#��2$ ��5Z�3�M�����˳�52����(c)...Ï��!�?DNr5�:�i�+ �"��#-M�{ O0�K��ts�2W��O-hk�|{�Ç�Luh��b1/4 ���d�S" �J_���H���4K�O��Ì�&9m;"��j���� �t�x2+�8�3��'�.p��>�*W���...K�,+�= �K�������!Xs��<y(� c���Q������e��괣...m��i�}.�N�H+�"�m^��x>R���&�O-ZM7&�uj���-M��Z�G>sÕª*'�'O�g��c��1�M"��$�m��m��_��� j �-!<i�2a-��?'�G��!"o0'�v�?@�n>��oxÛ��...*���(tm)�%�(tm)FF���-=d$��e.!'(c)�cÒ�'��y�_��3���e �O�z�í§]<���kX��v�@n�'��1/4�q��)C�5��8�sl"\$�~�,��?�po\�g�� �x�:Lg(r)$� ��)y�B$���F�5E���4@J$�n��?�p�Y���]G'Ë¥ckÛ@�8+4��C�Hñ��FD >���ADD"�-eN�!� 4v�W9�sQ�t� �...N�I E"X G3/4�'��g�e��.��>'��p-7mj7�'�=2�ج���GҰ�sÛa) Û¸o��EiG5�N...7...�p @���P�0�W"%�"n"RB'��"P�7�r�-�.b�@ p�$����#}���do\��w�lB�,�"� �*�ztk_�3��9��HO�6a�~.� -|{]�=�$dAV� ���HwjX��L�x�I�b�� �0��Z �'F�ʰ-�F��W6Þ��\����D�K�"�����o��ï¥G�?Q��M�..."�)�7��~5�G#�ɿ!Q'~.�fH �K����O��I"���q� ����9F�q��J��B����,?�a3/4E�%>#�DÈ·V~Q�%�@ ztEO�(-���� (r)q�p�[�'>)+�� �ۺ`~ "�mg�B��${���'���|Cd1/4'�46��"!3��oE.(tm)���~��3Ð"(c)�M��'8���"�1�� ^B-;"�à¯ï¿½1/4��3*I3�F��N�m�#�S7�b�j�E�!

  2. Math IB SL BMI Portfolio

    the data points in those areas, therefore the b value should be changed from ?/15 to ?/14. The number ?/14 was chosen because reducing the denominator's value by 1 will increase the b value by a slight amount, and an increased b value will shorten period and compress the graph

  1. Maths SL Portfolio - Parallels and Parallelograms

    3 3 9 18 30 45 63 4 6 18 36 60 90 126 5 10 30 60 100 150 210 6 15 45 90 150 225 7 21 63 126 210 etc. From the table, if m = 2 and n = 5, then p = 10.

  2. Logarithm Bases Math IA

    If the following two statements are true, Then this next statement must also be true, Therefore, the statements above will be used to prove that within this next statement: This investigation aimed to prove that this equation: is true for all similar logarithmic sequences.

  1. Artificial Intelligence & Math

    of arrests and decrease the amount of crime that is committed through the Internet. Hackers, cyber terrorists and white-collar criminals will be under threat by the system as their every keystroke and mouse click is monitored for clues to their criminal activities.

  2. Math IB HL math portfolio type I - polynomials

    = a + bi + c + di (2) (a + bi) - (c + di) =_a + bi - c + di (3) (a + bi) (c + di) = (a + bi) (c + di) (4) (a + bi)3 = (a + bi)3 _ _ If a is a real number, a = ? 0 = ?

  1. IB Math Methods SL: Internal Assessment on Gold Medal Heights

    function and the linear function (reproduced below): Information Table 2 (Table of values from linear equation y = 1.02x + 187) Years Elapsed (t) 0 4 8* 12* 16 20 24 28 32 36 40 44 48 Height in cm (h)

  2. SL Math IA: Fishing Rods

    Therefore we have determined that the quadratic equations given the points {(1,10), (6,96), (8,149)} is . Averaging of the Two Equations The next step in finding our quadratic function is to average out our established a, b, and c values from the two sets data.

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