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Logarithm Bases Portfolio

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Introduction

Math Portfolio Task 1: Logarithm BasesShyamal H Anadkat

Math Portfolio Task 1

Logarithm Bases

8/12/2012

IB-1 Mathematics

Shyamal H Anadkat


Introduction:

This portfolio is based on the mathematical concept of Logarithm Bases. The aim is to solve logarithm and exponential equations using log laws and base theorems and also to use and apply other mathematical concepts like series and sequences and exponents and relate them to logarithm concept. The portfolio is subdivided into three parts. Each part has a different question.

Part 1: The Task:

image12.png

Sequence 1 :

Term

Logarithm terms

Base

1

Log 2 8

2^1

2

Log 4 8

2^2

3

Log 8 8

2^3

4

Log 16 8

2^4

5

Log 32 8

2^5

6 (first next term)

Log 64 8

2^6

7 (second next term)

Log 128 8

2^7

->General: 2^n
Explanation for the next two terms in the sequence: Terms # 6 and 7 are the succeeding terms in this sequence. The argument being the same (8) in all the first five terms of the sequence there is a pattern observed in the bases which is the base of the first term is 2
1 , the second term 22 and hence according to the exponential pattern the base of the sixth and seventh term has to be 26 and 27 which is 64 and 128.

To find the nth term of this sequence in form of p/q:                                                                                    Log 2 8, Log 4 8, Log 8 8, Log 16 8, Log 32 8, Log 64 8, Log 128 8

The bases can be written as 21, 22, 23 up to 27. Hence the argument being same(8) bases can be generally denoted by 2n where n is the number of term.

...read more.

Middle

3

Log 125 25

4

Log 625 25

5 (new term)

Log 3125 25

6 (new Term)

Log 15625 25

For the 5th and the 6th term only the base changes as the Argument remains the same with the previous four terms in the series. The pattern followed by the bases is the same we have figured so far and in this case the general base formula is 5n. Hence for the 5th and the 6th terms have the bases 55 and 56(3125 and 15625) respectively and 25 as the same argument.

To find the nth term of this sequence in form of p/q:                                                                                    -Log 5 25, Log 25 25, Log 125 25, Log 625 25, Log 3125 25, Log 15625 25

The argument remaining the same i.e. 25, the bases in this sequence can be written generally in the form 5^n ( 51 for the first term, 52 for the second up till 56 for the sixth)

Hence the logarithm equation for the nth term is: Log 5^n 24

Let Log 5^n 25 = y, y= nth term in p/q form.

Then according to logarithm law:

5ny= 25  5ny= 52

ny= 2(according to exponents law) and y= 2/n

Therefore the expression is: Log 5^n 25 = 2 / n

Sequence 4:

Term

Logarithm terms

1

Log m m k

2

Log m2m k

3

Log m3m k

4

Log m4m k

5 (new term)

Log m5m k

6 (new Term)

Log m6m k

The 5th and 6th term are in sequence of the previous four with Argument mk

...read more.

Conclusion

Now using the log identity – Log b (mn) = Log b m + Log b n; log ab= log a + log b (base 10 or k)

Hence, log abx= image05.png

Substituting log a and log b from the 1 and 2 equations derived in the previous part:

           =Log abx= image06.png

       = Log abx=image07.png

(TAKING L.C.M AND MULTIPLYING)

           = Log abx=image08.png

(taking log x common)

         = Log abx= logimage09.png

  X   image10.png

         Hence,  Log abx= image11.png

is the general statement that expresses logabx in terms of c and d.

a= 4;         b= 2;       x=64

a=9;        b=3;        x=81

 Log 4 64 = 3, Log 2 64 = 6

 Log 9 81 = 2 , Log 3 81 = 4

Log a b X = C D / C + D

Log 4 ( 2 )64 = 3 ( 6 ) / 3 + 6

Log 8 64 = 18 / 9

Log 8 64 = 2

Log a b X = C D / C + D

Log 9 ( 3 ) 81 = 2 ( 4 ) / 2 + 4

Log 27 81 = 8 / 6

Log 27 81 = 4 / 3

Verifying :  Log 8 64 = 2

Verifying:  Log 27 81 = 4 / 3

8 2 = 64

64 = 64

27 ( 4 / 3 ) = 81

81 = 81

*Testing the validity of the general statement:

                               1                                                     2

Scope/limitations of a , b , and  x in this equation:

- In logabx= cd/c+d; x cannot be a negative number as no positive number put  to an exponent can give a negative result. Hence it is undefined.

- if x is zero; equation is undefined-logab0=cd/c+d ; as no power raised to a number would give zero as a result.

-if x=1; then logab 1= 0 (as anything raised to zero is one)

Conclusion:

To conclude the portfolio; the concept of logarithm laws and bases has been successfully applied to complete the tasks and proceed with the equations and to find the general statement.

Overall concept of logarithms and relating it with exponents has been carried out well.

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...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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