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

IB MATH TYPE I Portfolio - LOGARITHM BASES

Extracts from this document...

Introduction

LOGARITHM BASES

Purpose: My purpose for this investigation was to find the general statement that expresses logabx, in the terms of c and d. I was able to achieve this goal through the process of finding the expression for the nth term in various different sequences.

In the beginning stages of my investigation I came across the sequence of

        Log28, Log48, Log88, Log168, Log328.......

While I was looking at this sequence I came to the realization that the base of the log form was increasing at a constant pattern. I realized that the base of 2 was being raised to the power of the term number. Therefore the second term has a base of 4 due to the fact that 2 raised to the power of 2(term number) equals 4. Even though the base changes the 8 stays constant and this also means that I would be able to continue the pattern.

        6th term = log(2˄6)8                                  7th term = log(2˄7)8

                      = log648                                                    = log1288

After continuing the pattern I realized the next step would be to solve this problem and discover what exponent we need to raise to base to, in order to achieve the answer of 8. So I decided to change my pattern from log form into exponential form.

        2x=8,    4x=8,    8x=8,    16x=8,    32x=8,    64x=8,    128x=8

After successfully converting into exponential form I decided to take this a step further and make all the bases the same throughout the sequence.

        2x=23,22x=23,      23x=23,      24x=23,      2

...read more.

Middle

        =4                              =1

        34=81                            811=81

        81=81                            81=81

To further investigate expressions for different sequences will considered also consider these last sequences..

        Ex. log525, log2525, log12525, log62525, log312525, log1562525

        Ex. logmmk, log(m^2)mk, log(m˄3)mk, log(m˄4)mk, log(m˄5)mk, log(m˄6)mk  

Observing the previous sequences I will follow the same pattern and try to come up with and expression in the same form I have used for the pervious sequences.

        5x=25, 25x=25, 125x=25, 625x=25, 3125x=25, 15625x=25

        5nx=52, 5nx=52, 5nx=52, 5nx=52, 5nx=52, 5nx=52

        Therefore the expression is nx=2 or x=2/n    

Now I will check the validity of my expression with log2525, log62525

        x=2/n            x=2/n

        x=2/2            x=2/4

        x=1                x=0.5

Now I will check my answer

        Log2525         log62525

        =log25           =log25

          Log25            log625

        =1               =0.5

        251=25            6250.5=25

        25=25              25=25

        m1(x)=mk, m2(x)=mk, m3(x)=mk, m4(x)=mk, m5(x)=mk, m6(x)=mk

        mnx=mk, mnx=mk, mnx=mk, mnx=mk, mnx=mk, mnx=mk,

        Therefore the expression is nx=k or x=k/n    

Continuing my investigation I decided to look at other sequences however this time instead of

finding an expression that represents the sequence I will calculate the following sequences by giving my answer in p/q form.  

                                  Log464, log864, log3264

Log4

...read more.

Conclusion

ax = c, logbx = d, logabx .  This general statement does not work for all sequences for example it will work for the sequence of log28, log48, log88.  

Now that I have come up with a successful general statement I would like to recall how I was initially able to come up with a general statement. When I look back at the first set of example I did with the longer sequences it helped me understand the way an expression could be formed in the format of p/q. It just gave me an overall understanding of how logs work and how you can convert logs into exponential which can be beneficial when solving for logs. Later when I moved on to my second set of example with the set of three terms in each sequences really helped me with finding the final general statement. My previous experience of forming expressions in the p/q form really helped me with these sequences. When I solved each term in all the sequences and put the answers in fraction form it became very obvious how I could use the first and second term answer to arrive at the third term answer. Seeing all the term answers side by side really helped me connect all the dots together. I was able to tell if I multiplied the two answers then divided them by the sum of the two terms I will be able to arrive at the answer.  

...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

    c#1/2K rt���������M�`Jp���Z����'-��×U�%� -� �@XC`��W*2'��ÅP�-��ÈHdN$�����7�ɯy��1/4�*'�"|�7�_ ��y�M����]�Wp��+�gy�f�]�[����8Z��VF ��:hu��fAsi�Z ���Ek"c��.���=����P(tm)���� >����w�x�?{Yl_�g� �?)nÇ #�'c����'��/�g-�+#ŧ '/���n����}��� -���BRP�E�����O����i"�SE��]o��"ß´w0�7�C,"�A �1/4g 4�-0`pn`���`/�� '@8 j@=hW@;�}�6�F�0��,�-�7�A "C�;� C'���@Æd�A�P �C{�}P>TU@g��2� �Aw�G�3� ��@?`L3�ܰ, "���%�{Âp4��p\ 7�mp|~O��U@'P,(~"4J e��...rG bP�C�T5�"� u 5��F-���X4#�-��(c)� íF���+��6�z�1/2"�Â1\I��� �$br1%�:L+f�3����bY��XU�� �M�-�Va[�1/2�G��* �c�I�q"pT\.W�k��Ä�fq��$</^o�w�G��%�F|~ ?�� �" "]?B2�(�-�ExH�%l��Dm�#1"�E,#^$ _��H$':ÉL�$�'.'�HoH�4 C�x���41/24�h3/4��d�-ÙG. 7�o'_'��2��Ð��f�VÒ¶Ñ�~�#� ����K�+�"J��n'�@/BoHO�O����J��� ϰ�!��0C#�]�y �"B1��Qr(5"["F� �!�/�>�Z�A�Y&,"("9SS>S3��3...Y��(tm)9��'��4 �E"Å%��(�- -�ܬ����y�Y�X��8���� ���=a����n�-�~�1/2��% �C�Î#'�� �"'�&�/�!�+�Ϲ`.

  2. IB SL Math Portfolio- Logarithm Bases

    So, one could input 2x -- 8=0 and press solve to check their answer). The first sequence looked like this: log2^123 = 3/1 log2^223 = 3/2 log2^323 = 3/3 log2^423 = 3/4 log2^523 = 3/5 : : log2^n23 = 3/n The nth term of this sequence can be expressed as 3/n.

  1. Logarithm Bases - 3 sequences and their expression in the mth term has been ...

    U3: = log2781 (1.33) U4: : = log8181 (1) U5: : = log24381 (0.8) U6: : = log72981 (0.66) The graph above is a result of plotting f(x)= and plotting the values of each term in the sequence. (C) Now, the sequence of the 3rd row has the expression: Log5n 25 (this is because, the bases (b)

  2. Ib math HL portfolio parabola investigation

    would intersect the two lines at four distinct points and I have found these four points, labeling the intersections with y=Px+p as X2 and X3 and the intersections with y=Nx+n as X1 and X4. Using these four points I will again find the value of D = | (X2 - X1)

  1. Math Portfolio Type II Gold Medal heights

    The x-axis begins at 30(1926). One can see that the graph fits really well with (36, 197), (40, 203) and (84, 236). However the other points are not really included on the graph. Therefore it is fit to create a graph which suits the other points and then average the

  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. Gold Medal heights IB IA- score 15

    When t = 1940 h (1940) = 19.5 sin {/48 [(1940) ? 1962)]} +215.5 ?196.167 ?196 Therefore, the approximate gold medal height achieved in the year 1940 would be 196 cm. When t= 1944 h (1944) = 19.5 sin {/48 [(1944)

  2. High Jump Gold Medals Portfolio Type 2 Math

    The linear equation shown in Figure 3 and refined using Graphical Analysis fits the data well, however it is unrealistic to

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