# Math IA Logarithm bases

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Introduction

Logarithm Bases

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

Log 2 8 , Log 4 8 , Log 8 8 , Log 16 8 , Log 32 8 ,Log 64 8 , Log 128 8

Log 3 81 , Log 9 81 , Log 27 81 , Log 81 81 , Log 243 81 , Log 729 81

Log 5 25 , Log 25 25, Log 125 25 , Log 625 25, Log 3125 25 , Log 15625 25

Find an expression for the nth term of each sequence. Write your expressions in the form P/Q.

Log 2 8 , Log 4 8 , Log 8 8 , Log 16 8 , Log 32 8 ,Log 64 8 , Log 128 8

This sequence can be expressed 3/N. N being the nth term of the sequence.

1st Log 2 8 = 3 3/1 = 3

2nd Log 4 8 = 1.5 3/2 = 1.5

3rd Log 8 8 = 1 3/3 = 1

4th Log 16 8 = .75 3/4 = .75

5th Log 32 8 = .6 3/5 = .6

6th Log 64 8 = .5 3/6 = .5

7

Middle

This sequence can be expressed 4/N. N being the nth term of the sequence.

1st Log 3 81 = 4 4/1 = 1

2nd Log 9 81 = 2 4/2 = 2

3rd Log 27 81 = 1.33 4/3 = 1.33

4th Log 81 81 = 1 4/4 = 1

5th Log 243 81 = .8 4/5 = .8

6th Log 729 81 = .66 4/6 = .66

Find an expression for the nth term of each sequence. Write your expressions in the form P/Q.

Log 5 25 , Log 25 25, Log 125 25 , Log 625 25, Log 3125 25 , Log 15625 25

This sequence can be expressed 2/N.

Conclusion

(2)(-2)/(2 + (-2)) = -4/0

Using the general statement only further proves how (a)(b) cannot = 1 because it would cause you to divide by 0 which we are not able to do.

The general statement that I applied throughout the assessment (c)(d)/(c + d) I arrived at through plugging numbers in. Through a system of trial and error I was able to figure out the pattern and discover how you can deduce the third answer from the first two in the sequence using logs. Of course the general statement only applies to sequences of logs that follow the pattern of Log a X = c, Log b X = d , Log ab X = (c)(d)/(c + d).

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

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