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Logrithum bases

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Introduction

DaeYong Jang

Pd. 4th

Investigating Logarithm Bases

In this portfolio, I’ll attempt to investigate logarithm bases numerically and graphically by TI-83 plus graphic calculator, and other graphing soft wares. In part one to two, I’ll analyze the given logarithms and try to find an expression for the nth term of each sequence. In part three to five, I’ll use previous expression I found to find out how to get 3rd answer from first two answers, and create more examples from previous exercise. (Part I to V… The rest of them will be finished on due date.)

Part I

Here, I will analyze and try to write down next two terms of each sequence correctly. I will use TI-83 calculator to double check my work.

• Sequence 1: Log 2 8, Log 4 8, Log 8 8, Log 16 8, Log 32 8…
• Sequence 2: Log 3 81, Log 9 81, Log 27 81, Log 8181…
• Sequence 3: Log 5 25, Log 25 25, Log 125 25, Log 625 25…

:

• Sequence 4: Log m mk, Log m2mk, Log m3mk, Log m4p mk …

As I looked at the each sequences, I found out that base if the logarithm increased by number of power by that many terms from original base. This rule I found was proved by the last example of sequence 4.

Middle

p/q

6/1 (6)

6/3 (2)

6/2 (3)

Log 4 1024, Log 32 1024, Log 8 1024

 Logarithm 1st 2nd 3rd p/q 10/2 (5) 10/5 (2) 10/3

Log 7 343, Log 1/7 343, Log 1/49 343

 Logarithm 1st 2nd 3rd p/q 3/1 (3) 3/-1 (-3) 3/-2

Log 1/5 125, Log 1/25 125, Log 1/125 125

 Logarithm 1st 2nd 3rd p/q -3/1  (-3) -3/2 -3/3  (-1)

Log 2 512, Log 8 512, Log 4 512

 Logarithm 1st 2nd 3rd p/q 9/1  (9) 9/3  (3) 9/2

From the first table above, I noticed at first that you divide first two logarithms to get the third answer. If the first and second equation is divided, it will looks like:

Log 64 X Log 8                Log 64 X Log 8

Log 2 X Log 64                Log 2 X Log 64

The Log 64 cancel each other out and only Log 8/ Log 2 is left. If you put Log 8/ Log 2 into your calculator, the answer is 3, which is third logarithm’s answer in p/q form. However, rest of 4 sequences did not worked like the first sequence. I was able to see the clear connections between the bases of logarithms, but failed to found the relationship between 1st and 2nd to come up with the 3rd logarithm. I found out that bases are closely related, and was able to establish the pattern just for the base: 2nd base divided by 1st base will give 3rd logarithm’s base in all 5 sets. Also it had the same value m…

So then I substituted the 1st and 2nd logarithm as a and b respectively. I tried to multiply, subtract, add, and combination of these method. Finally, when I tried multiplication, division and subtraction all together, the fitting pattern appeared. The pattern I found is like this:

*a= 1stlogarithm, b= 2ndlogarithm*

ab       = 3rd logarithm

(a-b)

For example, I used 2nd set of logarithms, which is Log 4 1024, Log 32 1024, Log 8 1024. If I substitute first and second logarithm with a and b respectively, I will get following numbers:                      (5)(2)     =   10

(5-2)          3

Here, I can see that it gives me the 3rd answer in form of p/q form. So I ended up with two distinctive patterns. With this two patterns I found, I will now try to make same kind of sets see if it follows the rules when I place different numbers in the base.

Part V

Here in part five, I’ll try to create two more examples that fit the two patterns from Part IV within of my knowledge of understanding it. I’ll use calculator to get the answer that has fraction, and it is rational.

Example 1: Log 1/9 729, Log 1/729 729, Log 1/81 729

 Logarithm 1st 2nd 3rd p/q -3 -1 -3/2

Example 2:  Log 6 216 Log 216 216, Log 36 216

 Logarithm 1st 2nd 3rd p/q 3 1 1.5

Conclusion

Log m mk / nth. So In part 3 and 4, I tried to find a pattern that was in the given sets of logarithms. It took some time, but I found out that three logarithms had same x and 3rd Log’s base was product from 2nd Log’s base being divided by 1st Log’s base.

After discovery, I attempted to find a pattern for the set by substituting first two Logs with a and b combinations of adding, subtracting, division and multiplication. It took me a long time, but I was able to find the expression 3rd Log = ab /(a-b). Later in part six, I was able to confirm that the expression I found was the general expression I was looking for the whole time. After investigating further into the expression, I came to a conclusion that Log b/a X = cd/(c-d) is true. There is limitation to this expression, however. If a, b, or x ≠ (0, 1)U(1, ).

As I finished investigating the bases of common logarithms, I came to a conclusion that you can get third logarithm’s answer from the two Logs if I use expression Log b/a X = cd/(c-d) if a, b, and x= (0, 1)U(1, ∞), and x has to be constant in both Logs.

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