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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 , …

:

:

:

, ,,, …

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 , ,

Log381 , Log981 , Log2781 , Log8181 , ,

Log525 , Log2525 , Log12525 , Log62525 , , ,

, ,,,,

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

 1 2 3 4 5 6 7 2 2 2 2 2 2 2 2 4 8 16 32 64 128

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

Middle

27

81

243

729

2187

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

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

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

Here are two graphs to test its validity and accuracy:

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 , , ,) into consideration:

 1 2 3 4 5 6 7 5 5 5 5 5 5 5 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     by applying the change of base rule therefore:

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

Here are two graphs to test its validity and accurateness:

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 (  ,, , , , ) into concern:

 1 2 3 4 5 6 m m m m m m

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:

=

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:

=

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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