# Logarithm Bases - 3 sequences and their expression in the mth term has been given. All of these equations will be evaluated on a step-by-step basis in order to find an expression for the nth term.

Extracts from this document...

Introduction

Mathematics Internal Assessment

## LOGARITHM BASES

Abhishek Puri

IB Mathematics-SL

29/05/2009

Mr. Adam Scharfenberger

Part- 1

3 sequences and their expression in the mth term has been given. All of these equations will be evaluated on a step-by-step basis in order to find an expression for the nth term. The technology which will be used to execute this investigation will be a TI-83 graphing calculator and Logger Pro graphing Program. All mathematical calculations have been made by this graphing calculator and is mentioned where some calculations have not been shown.

(A)

Let us observe the first row of the sequence.

The following sequence is given as:

Each of the terms in this sequence are of the form . If analyzed, it can be easily observed that the log base (b) of each term in this sequence is a power of 2 while A remains constant at 8.

Therefore, we can state that each term for the 1st row has the general expression log2n 8, where n is the nth term of the sequence. Now, we must apply this expression in the form .

log2n 8

= [Applying change of base formula]

= [Substituting 8 as a power of 2]

=

Now, the expression , should provide us with the values of each term in the sequence. Thus, to prove as the general expression for the first row of sequences:

un =

First term of the sequence =

= log2n 8

(Taking the first term as u1 and so on,)

L.H.S R.H.S

u1 = log21 8

u1 = 3 log28 = 3

L.H.S = R.H.S. Hence, verified that is the general expression for the 1

Middle

The graph above is a result of plotting f(x)= and plotting the values of each term in the sequence.

(C)

Now, the sequence of the 3rd row has the expression:

Log5n 25 (this is because, the bases (b) of each of the terms in the sequence are powers of 5 while A, which is 25,remains the same for each term)

= [Applying change of base formula]

= [Substituting 25 as a power of 5]

=

Now, the expression , should provide us with the values of each term in the sequence. Thus, to prove as the expression for the second row of sequences in the form :

un =

First term of the sequence =

= log5n 25

(Taking the first term as u1 and so on,)

L.H.S R.H.S

u1 = log51 25

u1 = 2 log525 = 2

Similarly, other terms in the sequence were also verified by using the TI-83 where:

U2: log25 25 (1)

U3: = log12525 (0.66)

U4: : = log62525 (0.5)

U5: : = log312525 (0.4)

U6: : = log1562525 (0.33)

L.H.S = R.H.S. Hence, Verified that is the general expression for the 3rd row of the given sequence in the form .

(D)

Also given, is the sequence for the mth term:

Logm mk, logm2 mk , logm3 mk, logm4mk, …

Thus, we can establish the expression of this sequence to be:

Logmn mk

=

=

Therefore, the general expression (which is in the form ) can be applied to all the given sequences. We can apply this general expression to verify the successive terms of each sequence.

Term (u) | Logarithm of 8 with base 2n | Logarithm of 81 with base 3n | Logarithm of 25 with base 5n |

1 | 3.00 | 4.00 | 2.00 |

2 | 1.50 | 2.00 | 1.00 |

3 | 1.00 | 1.33 | 0.66 |

4 | 0.75 | 1.00 | 0.50 |

5 | 0.60 | 0.8 | 0.40 |

6 | 0.50 | 0.66 | 0.33 |

7 | 0.42 | - | - |

Conclusion

Similarly, let us get the values of the logarithms in the first column. We obtain 3(log464), 2(log749),-3(Log 125) and 3 (log8512) .

Now, we have found out the all values of c and d for in the sequences given to us. The grid looks like:

Column 1(c) Column 2(d)

3 2

2 1

-3 -1

3 9

However, we can now get back to our formula we can now add values to the third column by applying this formula to each of the values in the first and second columns respectively .(with the help of the graphing calculator)

Therefore, Column 3

1.5

0.66

-.075

-4.5

Thus, by verifying with a graphing calculator, these values above are equal to the values of the logarithmic terms given in the sequences.

Therefore, following the pattern, another example which will fit the pattern above is.

Log6 36, log36 36, log 216 36

Log 729, log729 729, log 81 729

This is proved by :

Log6 36= 2

log36 36= 1

log 216 36=

according to the formula we get

Hence, since is equal to the value of the last term, the first sequence is verified.

For the second sequence,

We obtain Log 729 = -3

log729 729 = 1

log 81 729 = 1.5

according to the formula we get or 1.5

Hence, since is equal to the value of the last term, the second sequence is also verified.

The limitations of a, b and x is that we can only take positive values of a and b because logarithms of negative values don’t exist. Therfore, a, b and x can’t be negatives even though it is given that p,q, Therfore, p,q are only confined to p,q[+]

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

## Found what you're looking for?

- Start learning 29% faster today
- 150,000+ documents available
- Just £6.99 a month