Logarithm Base
IB Math SL
Type I Portfolio
Lisa Phommaseng
Logarithm Base
Consider the given sequences:
Log28, log48, log88, log 168, log328,…
Log381, log981, log 2781, log8181,…
Log525, log2525, log12525, log62525,…
:
:
:
,…
For the first sequence Log28, log48, log88, log 168, log328,… you are to find the next two terms of each of the sequences you would have to determine the pattern. As we can see, the value of the bases, 2,4,8,16,32 are increasing by squares of 2 (ex: 22= 4, 23=8, 24=16 etc.,). The two terms following log328 will be log648 and log1288. To find the expression for the nth term of each sequence, you would have to write the expression in the form of . First you would find log28 to begin to look for a pattern. A formula that could be useful to solve this would be logbx = . The variable b would be the base, x would be the number of the logarithm. Saying that log28 = x, we are able to determine that x is 3 because 23 equals 8. The same would follow for the next terms : log48= x, 8 is equivalent to 23 and the base of 4 is equivalent to 22 (Log28 =). So, to that term would be . You would continue on with the next and you would notice that the pattern is 3 over whatever the number that follows. Now we see that the nth term of this sequence is an=. The next sequence Log381, log981, log 2781, log8181,…you would determine the following two terms as the one above. A we can see the terms are squaring as well. The bases are increasing as so : 31=3, 32=9, 33= 27, 34=81, and so forth. We can determine that the following terms are going to be log24381 and log72981. You would also find the expression for the nth term the same way. Using the formula logbx = we are able to see that log381 = x and x = 34 or 4. Following that log981 is equal to for log981 = . The nth term of this sequence is an=. So, the following two terms of the last two sequences would be log312525 and log1562525 for the sequence Log525, log2525, log12525, log62525,…...
...and b, I use the method of solving matrix which is:
Thus,
We plot the graph with new function as follows:
Then we compare new function with two curves mentioned above in one graph as follows:
From the graph, it can be concluded that the new function can fit the data, and better than the first function, but there is still some error between these two curves. New solution method or model should be advanced to describe tolerance of human beings to Gforces over time.
Then, we still found that some little error exist between real data and the function we got. Observe the time of the data, it is clear that there are large difference between the data, the scale of time is from 0.01 to 30, which is that uniformly distribute. So, logarithm axis is considered to scale the time. We can get new data table as follows:
Time (min)  Log Time (min)  Ln Time (min)  +Gx (g) 
0.01  2  4.61  35 
0.03  1.52  3.51  28 
0.1  1  2.3  20 
0.3  0.52  1.2  15 
1  0  0  11 
3  0.478  1.1  9 
10  1  2.3  6 
30  1.478  3.4  4.5 
Then we plot the graph of Log Time and Ln Time as follows:
From the graph, it can be seen that the shape of curve is similar with liner function. We assume the function form as follows:
So, we still use the new data to solve parameter k and b, we choose first and fifth group of data, second and the sixth group of data, third and seventh group of data, fourth and eighth group...
...Taipei European SchoolMath Portfolio

VINCENT CHEN 
Gold Medal Heights
Aim: To consider the winning height for the men’s high jump in the Olympic games
Years  1932  1936  1948  1952  1956  1960  1964  1968  1972  1976  1980 
Height (cm)  197  203  198  204  212  216  218  224  223  225  236 
Height (cm)
Height (cm)
As shown from the table above, showing the height achieved by the gold medalists at various Olympic games, the Olympic games were not held in 1940 and 1944 due to World War II.
Year (1=1932, 2 = 1936 and so on)
Year (1=1932, 2 = 1936 and so on)
Using autograph, the graph above is a scatter graph showing the high jump results from the table.
The plot suggests that the high jump heights start off with a steep positive slope then coming to a decreasing negative slope, however without the 1940 and 1944 high jump competitions, it may not be certain. Then finally, it starts increasing again with a fairly steep positive slope.
However, it would not be realistic if the function has an infinitely increasing range, such as quadratic, exponential and linear because of the limitations that humans have due to natural forces like gravity. Therefore, narrowing down the options that may fit this graph to natural logarithm and logistics
Since the statistics given starts from year 1896, in order to make sure that calculations can be as simplified as possible, I have decided to rearrange the...
...MATHS PORTFORLIO SL TYPE I
CIRCLES
In this portfolio I am investigating the positions of points in intersecting circles. (These are shown on the following page.
The following diagram shows a circle C1 with centre O and radius r, and any point P.
The circle C2 has centre P and radius OP. Let A be one of the points of intersection of C1 and C2. Circle C3 has the centre A, and radius r. The point Pʹ is the intersection of C3 with (OP). This is shown in the diagram below.
As shown on the assignment sheet, r=OA. We therefore need to find the values of OPʹ when r=OA=1 for the following of the values of OP: OP=2, OP=3 and OP=4.
We first of all extract the triangle OPA from the above diagram and since we have the values for all the three sides we can finds the angle AOP which will later on help to get the value of length OPʹ. The circle C2 and triangle OPA are shown below with all side of OPA indicated.
OP=AP since they are the radii of the same circle, C2. Having all the three sides, we can now calubulate the angle AOP using the cosine rule. Angle AOP is calculated below:
Cos AOP=(2^(2 _ ) 2^(2 _ ) 1^2)/(2×2×1)
Cos AOP=0.25
∴ AOP =COS10.25
=75.52248781
≈75.5˚...
...Exploration of Lacsap’s Fractions
The following will be an investigation of Lacsap’s Fractions, that is, a set of numbers that are presented in a symmetrical pattern. It is an interesting point that ‘Lacsap’ is ‘Pascal’ backwards, which hints that the triangle below will be similar to “Pascal’s Triangle”.
1 1
1 1
1 1
1 1
1 1
There are many patterns evident in this triangle, for instance I can see that there is a vertical axis of symmetry down the middle of the triangle. Each row starts and ends with the number 1. Each row has one more variable than the number of rows, i.e. row 1 has 2 variables. The numerators in the middle stay the same and the diagonals form sequences.
In order to decipher the pattern in the numerators and denominators, I had to look at the triangle a different way. Knowing that the numerators of the row don’t change, it occurred to me that the number 1s on the outside of the triangle could be expressed as fractions.
This proves that all the numerators of the row are the same.
To further investigate the numerators, I will examine the relationship between the
row number and the numerator, which is shown in the table below. These are the numerators after having changed the 1s on the outside of the triangle to their fraction forms, thus...
...Math SL Portfolio – Tips and Reminders Checklist
Notation and Terminology
Check for the following:
• I did not use calculator notation. (I didn’t include things like ‘x^2’ for or Sn for Sn)
• I used appropriate mathematical vocabulary.
Communication
Check for the following:
• The reader will not need to refer to the list of questions in order to understand my work.
• My responses are not numbered.
• I have an introduction, conclusion, title page, and table of contents.
• All graphs are labeled – Each graph has a title, labeled axes, and appropriate scale.
• My graphs and tables are within the body of my work. They are not separate or in an appendix.
• I have explained why I made the choices I did when going through the task.
• I did not include key stroke sequences, e.g. “I pressed the 2nd key, then TRACE…”
• My tables do not straddle pages.
• My tables are labeled well, including my variable definitions in each column.
Use of Technology
Check for the following:
• I used technology to illustrate my points and ideas. I didn’t just “stick in” a graph.
• Each graph or table (or other piece of tech.) is accompanied by explanations and my ideas.
• I did not include too many graphs on the same axes – my graphs are easy to read.
Mathematical Process (Type 1)
Check for the following:
• I explicitly defined variables and parameters the first time I used them, even if they were
already defined in...
...Introduction: In this following assignment, I will be considering geometric shapes that lead to special numbers. The simplest examples of these are square numbers (1, 4, 9, 16, etc), which are derived from squaring 1, 2, 3, and 4. From this I got the equation y= x2. This equation is illustrated in the table below.
y=x2
x y 
1 1 
2 4 
3 9 
4 16 
In the table on the left, I observe that from the y value 1 to the y value 4 there is an increase of 3. From the y values 4 to 9, there is an increase of 5. From the y values 9 to 16, there is an increase of 7. This shows that it goes: +3, +5, +7, which is then increasing by 2 between each of those numbers.
Below, is the graph of y=x2
[pic]
The equation y=x2 comes from the general equation y= ax2bx+c.
Y=x2 is the same as y=x2+0x+0. Therefore, a=1, b=0, and c=0.
The next example I am going to show you is similar to the one above. The following diagrams show a triangular pattern of evenly spaced dots. The numbers of dots in each diagram are examples of triangular numbers (1, 3, 6, 10, 15, 21, 28, 36).
Tn y 
1 1 
2 3...
...Leonardo Souilljee Alberton
Personal Statement
I am a passionate, persistent individual who wills to become a bachelor in British Law, in order to, further on, exert the profession of a lawyer. I am and have always been deeply interested in the principles of justice, which I try to put into action in my everyday life.
My intention on entering a Law School is acquiring a deeper knowledge on the subject which has always been my main interest, both in my social and in my personal life. One might perhaps marvel that a Brazilian would rather join an British law school rather than one in his home country; the main reason why I am convinced that studying law in the United Kingdom, more precisely in England, is the absolute passion and interest I have always had for this country throughout my whole life, also extending to the field of law.
My interest in Law and Rights first blossomed at the age of seven, when problems like alimony and share of goods showed up after my parents’ divorce, raising my interest to the mutual dissatisfaction of each with different matters concerning law, which made me wonder to what extent Divorce Law in Brazil was fair to both sides of the divorce. Moreover, I also have been told the basics of the profession by uncles  who are bachelors in Brazil – and about how timedemanding and complex it can be. My uncle also told me, though, that as long as it is exerted for passion, and not for other reasons, the high charge of work and...
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