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Type I - Logarithm Bases

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Introduction

IB Standard Portfolio Assignment

Type I – Mathematical Investigation

Logarithm Bases

This investigation will determine the relation between different sets of sequences. The sequences include logarithms. This investigation will be tested using technology.

The sets of sequences are as follows:

Log28 , Log48 , Log88 , Log168 , Log328 , …

Log381 , Log981 , Log2781 , Log8181 , …

Log525 , Log2525 , Log12525 , Log62525 , …

:

:

:

image00.pngimage00.png, image15.pngimage15.png,image17.pngimage17.png,image18.pngimage18.png, …

By following these sequences a pattern can be shown. The base of each term in the sequences changes but the exponents are constant. The following two terms of each sequence were determined:

 Log28 , Log48 , Log88 , Log168 , Log328 , image88.pngimage88.png, image01.pngimage01.png

Log381 , Log981 , Log2781 , Log8181 , image16.pngimage16.png,image30.pngimage30.png

Log525 , Log2525 , Log12525 , Log62525 , image38.pngimage38.png, image48.pngimage48.png,image58.pngimage58.png

image00.pngimage00.png, image15.png,image17.pngimage17.png,image18.pngimage18.png,image19.pngimage19.png,image20.pngimage20.png

Let’s start with the first sequence (Log28 , Log48 , Log88 , Log168 , Log328 , image88.pngimage88.png, image01.pngimage01.png ) and determine an expression for the nth term:  

1

2

3

4

5

6

7

2

2

2

2

2

2

2

image89.png

image90.png

image91.png

image92.png

image93.png

image94.png

image95.png

2

4

8

16

32

64

128

The value 2 was used to determine the nth term.   image96.png  is worked out from the table above in the form of   image07.png

...read more.

Middle

27

81

243

729

2187

I used the value 3 to determine the nth term since the value of 2 did not work.   image114.png

is worked out from the table above in the form of   image07.png  by applying the change of base rule therefore:

image08.pngimage08.pngimage115.png

image08.pngimage08.pngimage116.png

image117.png

The numerator and denominator both have the value log3 in them which means they can be canceled out thus leaving us withimage118.png

Here are two graphs to test its validity and accuracy:

image119.jpg

image120.jpg

Both graphs are identical towards each other which indicate that both functions of the graphs are the same.

Let’s take the third sequence (Log525 , Log2525 , Log12525 , Log62525 , image38.pngimage38.png, image48.pngimage48.png,image58.pngimage58.png) into consideration:

1

2

3

4

5

6

7

5

5

5

5

5

5

5

image121.png

image122.png

image123.png

image02.png

image03.png

image04.png

image05.png

5

25

125

625

3125

15625

78125

I used the value 5 to determine the nth term.                 is worked out from the table above in the form of   image07.png  by applying the change of base rule therefore:image06.png

image08.pngimage08.pngimage09.png

image08.pngimage08.pngimage10.png

image11.png

The numerator and denominator both have the value log5 in them which means they can be canceled out thus leaving us with image12.png

Here are two graphs to test its validity and accurateness:

image13.jpg

image14.jpg

Both graphs are identical towards each other which indicate that both functions of the graphs are the same.

Let’s take the fourth and final sequence ( image00.pngimage00.png ,image15.pngimage15.png, image17.pngimage17.png, image18.pngimage18.png, image19.png,image20.pngimage20.png ) into concern:

1

2

3

4

5

6

m

m

m

m

m

m

image21.png

image22.png

image23.png

image24.png

image25.png

image26.png

...read more.

Conclusion

Clarification of How I Arrived at My General Statement:

The sequences given were firstly observed. I determined and have assured that each sequence had a constant exponent but the bases of each one were not. After observing the sequences I then switched the exponents and bases together with the help of the change of base rule. Basically I inversed them:

image78.pngimage78.png=image79.pngimage79.png

After observing the sequences and determining the nth term we had to test the validity of our general statement using other values of a, b, and x. The validity was proved by finding this formula:

image86.pngimage86.png= image87.pngimage87.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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