- Level: International Baccalaureate
- Subject: Maths
- Word count: 2113
Logarithm Bases
Extracts from this document...
Introduction
Math portfolio
Find the nth term of a logarithmic sequence
In this portfolio I will investigate how we can use changes in the base of a logarithmic sequence to find a general expression. We start with the fallowing logarithmic sequences;
:
:
:
We can use the last sequence of the sequences above to show a pattern of the change of base in the sequences. By using this information we can deduce the nth term of any logarithmic sequence. We can also see that the last sequence is written with the base and the number if the logarithm both using m raised to a power, but the number of the logarithm, mk, is the constant in the expression. If we look at the pattern of the other sequences we can see that their number is also constant. The m exponent increases proportionally to the term for each term it is an increase by 1.
Now that we have m both as a base and as the number of the logarithm we can use the change of base rule to simplify the terms:
Using the rules for logarithms we get the expression:
The use of m both in the base and the number of the logarithm is a smart thing to do. Because the
Middle
2
1
2
25
25
1
2
2
1
25
125
0,666666667
2
3
0,66666667
25
625
0,5
2
4
0,5
25
3125
0,4
2
5
0,4
25
15625
0,333333333
2
6
0,33333333
25
78125
0,285714286
2
7
0,28571429
25
390625
0,25
2
8
0,25
25
1953125
0,222222222
2
9
0,22222222
25
9765625
0,2
2
10
0,2
As you can see the solutions to the term in both forms are exactly the same, in other words, what we have done, writing the terms in the form , is a correct way of writing the term, because, as said the solutions are identical. As you can see I have continued the sequence to the tenth term, showing that it is correct not only for the calculations shown earlier. The formulas used in these excel sheets can be viewed in attachments 1 and 2 at the back of the portfolio. Let’s move on to the next section.
Applying to another pattern and its use
Now let’s try using some of our newly acquired knowledge to solve a problem involving logarithmic sequences similar to those we have just dealt with. Take a look at these four sequences:
If you look closely at these sequences you will notice that the terms are consecutive, in the three first sequences certain terms have been skipped, and in the last sequence the two first terms have been switched around. Even though these sequences do not seem to have much in common, there is more than meets the eye. It is possible to find the third answer the two first. To get an overview let’s rewrite the terms to the form .
First sequence:
Conclusion
Limitations:
The base must be positive and not equal to one, so:
a > 0, b > 0 (because each of them also shows up alone on a logarithm, they can't both be negative).
Also, ab ≠ 1 -> a ≠ 1/b
Finally, the argument (x) must be positive, so:
x > 0
And we want
So I'm going to rewrite (1) and (2) so that I can make appear
Therefore :
Now we have found a general statement for the expression logabx in terms of c and d. Where a >0, b>0, ab ≠ 1, c ≠ -d. We can therefore say that the general statement is .
A test of the general rule:
A = 1, B = 2, X = 2
log1×2 2 = ((log 2 ÷ log 1) × (log 2÷ log 2)) ÷ ((log 2 ÷ log 1) + (log 2 ÷ log 2))
1 = ((log 2 ÷ 0) × (1)) × ((log 2 ÷ 0) + (1))
1 = (0 × 1) × (0 + 1)
1 = 0 × 1
A can not be 1.
Attachment 1;
number of logarithm | base | log | k | n | k/n | |
8 | =2^1 | =LOG(B16;C16) | 3 | 1 | =F16/G16 | |
8 | =2^2 | =LOG(B17;C17) | 3 | 2 | =F17/G17 | |
8 | =2^3 | =LOG(B18;C18) | 3 | 3 | =F18/G18 | |
8 | =2^4 | =LOG(B19;C19) | 3 | 4 | =F19/G19 | |
8 | =2^5 | =LOG(B20;C20) | 3 | 5 | =F20/G20 | |
8 | =2^6 | =LOG(B21;C21) | 3 | 6 | =F21/G21 | |
8 | =2^7 | =LOG(B22;C22) | 3 | 7 | =F22/G22 | |
8 | =2^8 | =LOG(B23;C23) | 3 | 8 | =F23/G23 | |
8 | =2^9 | =LOG(B24;C24) | 3 | 9 | =F24/G24 | |
8 | =2^10 | =LOG(B25;C25) | 3 | 10 | =F25/G25 | |
number of logarithm | base | log | k | n | k/n | |
25 | =5^1 | =LOG(B3;C3) | 2 | 1 | =F3/G3 | |
25 | =5^2 | =LOG(B4;C4) | 2 | 2 | =F4/G4 | |
25 | =5^3 | =LOG(B5;C5) | 2 | 3 | =F5/G5 | |
25 | =5^4 | =LOG(B6;C6) | 2 | 4 | =F6/G6 | |
25 | =5^5 | =LOG(B7;C7) | 2 | 5 | =F7/G7 | |
25 | =5^6 | =LOG(B8;C8) | 2 | 6 | =F8/G8 | |
25 | =5^7 | =LOG(B9;C9) | 2 | 7 | =F9/G9 | |
25 | =5^8 | =LOG(B10;C10) | 2 | 8 | =F10/G10 | |
25 | =5^9 | =LOG(B11;C11) | 2 | 9 | =F11/G11 | |
25 | =5^10 | =LOG(B12;C12) | 2 | 10 | =F12/G12 |
Attachment 2;
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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