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IB Math Portfolio- Topic 1

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Introduction

International Baccalaureate

Math Methods SL Portfolio 1:

Logarithms

December 02, 2009

The intent of this portfolio is to explore, investigate, and model the patterns found in logarithmic functions.

In considering the first logarithm in the first sequence:

image05.pngimage05.png

It is important to know what it represents.

By definition, a logarithm is the inverse of an exponential function.

Therefore, when:

image69.pngimage69.png

image93.pngimage93.png

The next expression in the series is:

image77.pngimage77.png

By definition:

image120.pngimage120.png

This expression is more difficult to solve as easily.

However, if 8 is seen instead as:

image06.pngimage06.png

The expression becomes easier to solve

image27.pngimage27.png

image41.pngimage41.png

image50.pngimage50.png

image60.pngimage60.png

There is a pattern in the bases of each consecutive expression.

In the first set, the base is defined as:

image70.pngimage70.png

where n is the nth term in the sequence.

In considering the whole sequence:

Sequence 1:

Term 1

Term 2

Term 3

Term 4

Term 5

image05.png

image77.png

image78.png

image79.png

image80.png

So, in term 5, the base is image81.pngimage81.png

Therefore, in terms 6 and 7 of all sequences, the value n must be equal to 6 and 7 respectively

Term 6

Term 7

image82.png

image83.png

The graph below represents the resulting values in sequence 1 as the exponent of the base increases.

image84.jpg

...read more.

Middle

image112.png

image113.png

image114.png

The numerator stays at k, and since the base of the exponent is represented by n, the denominator is also represented by n.

Therefore, the image115.pngimage115.pngterm of the sequence in terms of p/q is image116.pngimage116.png

It is important to look at finding the solution to the answer in a different way.

The change of base rule is generally a rule that puts the logarithmic statement into one that is expressed in base 10, so it is easily calculatable on the calculator.

The change of base rule is this:

For image117.pngimage117.png

image118.png

This can easily be proved by considering the solution by definition of a logarithm.

image119.png

Now, calculate the following, giving your answers in the form image121.pngimage121.png

Sequence 2a

image11.png

image12.png

image13.png

Results

image122.png

2

1.2

Sequence 2b

image17.png

image18.png

image19.png

Results

image123.png

image124.png

image125.png

Sequence 2c

image23.png

image24.png

image25.png

Results

image126.png

image127.png

image128.png

Sequence 2d

image30.png

image31.png

image07.png

Results

image08.png

image09.png

image10.png

Figuring out the third expression in the sequence is a matter of adding together the exponents of the base.

In the first example, the bases have a similar root-- namely, 2.

Sequence 2a

image11.png

image12.png

image13.png

Think of the bases not as separate values, but instead as exponents of 2.

Sequence 2a(Revised)

image14.png

image15.png

image16.png

From the results above, it is easy to suspect that the third term of the sequence comes from an addition of the exponents of the bases in the two previous terms:

2+3=5

The same can be done with the next three sequences:

Sequence 2b

image17.png

image18.png

image19.png

Sequence 2b(Revised)

image20.png

image21.png

image22.png

From the results above, it is seen that the exponent of the third term is the sum of the two base exponents in the terms preceding it.

1+2=3

Sequence 2c

image23.png

image24.png

image25.png

Sequence 2c(Revised)

image26.png

image28.png

image29.png

(-1)+(-3)=(-4)

Sequence 2d

image30.png

image31.png

image07.png

Sequence 2d(Revised

image32.png

image33.png

image34.png

...read more.

Conclusion

image72.png

Then, by taking the inverse once more, the result for the general statement can be reached:

image73.pngimage73.png= image61.pngimage61.png

When solving for the general statement, and testing for values of a, b, and x, using the change of base formula caused some problems:

image74.png

By this thought, both x and a cannot be negative, because 10 to the power of any real number equal to a negative number.  image75.jpg

The same can be said for b:

image76.png

x and b cannot be negative.

In graphing the two equations, the asymptotes and limits can be seen as to where the values do not exist.

image04.png

It is seen in the graph that the x values of the graph are limited to positive values over 0.

in replacing x with 0, it is seen that the value cannot exist, because no  integer a or b, (a or b > 0) to the power of another integer (over 0) can equal a value of 0, but can approach it very closely.

The result for the general statement arrives from different methods.  In using the change of base formula as well as using a proven inverse rule, adding together two logarithms of different bases is possible.

...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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