Figure 7.
Figure 6.
Can you take the logarithm of square roots? View figure 8 below.
Figure 8.
Now, there are two basic logarithms common and natural. Beginning with common, the 10 of a number. That is, the of 10 necessary to equal a given number. The common logarithm of x is written log x. For example, log 100 is 2 since 102 = 100. The base is understood to be ten. View the diagram in figure 9. The natural logarithm is composed of the e of a number. The system of natural logarithms has the number called "e" as its base. Why e? (e is named after the 18th century Swiss mathematician, Leonhard .) It is called the "natural" base because of certain technical considerations. That is, the of necessary to equal a given number. The natural logarithm of x is written ln x. For example, ln 8 is 2.0794415... since e2.0794415... = 8. How is e calculated? View the diagram in figure 10. ln is the symbol for natural log.
Figure 9.
Figure 10.
There are three laws of logarithms:
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"The logarithm of a product is equal to the sum
of the logarithms of each factor."
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logbxy = logbx + logby
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"The logarithm of a quotient is equal to the logarithm of the numerator
minus the logarithm of the denominator."
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logb x
y = logbx − logby
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"The logarithm of a power of x is equal to the exponent of that power
times the logarithm of x."
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logb xn = n logbx
The laws of logarithms will be valid for any base. I will prove this using base e, that is, for
y = ln x. I am going to prove law number 1. "The logarithm of a product is equal to the sum
of the logarithms of each factor."
View the diagram in figure 11, below.
Figure 11.
The second law can be proved in the same way. Instead of addition in the steps it would be subtraction. View the example in Figure 12 to see an example of the first three laws in a problem.
Figure 12.
I will prove the third law, "The logarithm of a power of x is equal to the exponent of that power
times the logarithm of x." View Figure 13, below.
Figure 13.
As you know not all logs will hold a natural or common logarithm base such as 10 or e.
Therefore, change of base is necessary. View the diagram in Figure 14.
Figure 14.
What is another example of the change of base? View figure 15.
Figure 15.
The “base”ics
From the as inverse of an exponential, you can immediately get some basic facts. For instance, if you graph y=10x (or the exponential with any other positive base), you see that its range is positive reals; therefore the domain of y=log x (to any base) is the positive reals. In other words, you can’t take log 0 or log of a negative number unless you want to deal with imaginary numbers. View the diagram in figure 16 to see the graph of y=10x.
Figure 16.
A less congested version of all the rules and a few more can be viewed in the table in figure 17, below.
Figure 17.
There you have it! The general idea of what are logarithms. The puzzle pieces have been put together. Now onto the exploration of logarithm bases!
Problem 1:
Consider the following sequences. Write down the next two terms of each sequence.
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log 2 8, log 4 8, log 8 8, log 16 8, log 32 8,… The next two terms are log 64 8 and log 128 8
General expression for the nth term of this sequence is log2n 8 = 3/n
View the diagram below to see how the work is done. I saw a pattern with 2n and wrote down the relationship for the remaing sequence. The two boxes are interconnected as one side of work goes along with the other. Following, the logarithm rules of multiplication, division and powers and change of base.
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log 3 81, log 9 81, log 27 81, log 81 81,… The next two terms are log 243 81 and log 729 81
General expression for the nth term of this sequence is log 3n 81 = 4/n. View the work below to understand.
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log 5 25, log 25 25, log 125 25, log 625 25,… The next two terms are log 3125 25 and log 15625 25
General expression for the nth term of this sequence is log 5n 25 = 2/n. View the work below to understand.
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log m mk , log m2 mk , log m3 mk , log m4 mk ,...The next two terms are log m5 mk and log m6 mk
General expression for the nth term of this sequence is log mn mk = k/n. View the work below to understand.
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I found what the following terms were by following the patterns previous to the continuation of the sequence. By using GDC with the TI – 83 Plus I calculated the answers such as 56 for the third sequence.
Part 2 of Problem 1:
This involves the changing of the general expressions into the form p/q, where p,q .
To do this you are solving each logarithm.
To review the mathematics involved view figure 1.
1.) The first sequence:
The general expression for the nth term of this sequence is log2n 8 = 3/n.
Therefore the expressions will be: 3/1, 3/2, 3/3, 3/4, 3/5,... and so forth.
2.) The second sequence:
The general expression for the nth term of this sequence is log 3n 81 = 4/n.
Therefore, the expressions will be: 4/1, 4/2, 4/3, 4/4, ... and so forth.
3.) The third sequence:
The general expression for the nth term of this sequence is log 5n 25 = 2/n.
Therefore, the expressions will be 2/1, 2/2, 2/3, 2/4, .....and so forth.
4.) The final sequence:
The general expression for the nth term of this sequence is log mn mk = k/n.
Therefore, the expressions will be k/1, k/2, k/3… and so forth.
- To justify my answers with technology I used GDC with the TI – 83 Plus.
- I converted certain decimals to fractions using the calculator.
Problem 2: Now calculate the following giving your answers in the form p/q where p,q .
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log 464 = 3/1 log 864 = 2/1 log 32 64 = 6/5
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log 7 49 = 2/1 log 49 49 = 1/1 log 343 49 = 2/3
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log 1/5 125 = -3/1 log 1/125 125 = -1/1 log 1/625 125 = -3/4
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log 8 512 = 3/1 log 2 512 = 9/1 log 16 512 = 9/4
Part 2 of Problem 2:
Describe how to obtain the third answer in each row from the first two answers.
If log ax = c and log b x = d
therefore, the general statement that expresses log ab x in terms of c and d is:
log ab x = (cd)/(c + d) I arrived to this statement by combing all of the laws together.
Restrictions: a >0, b>0, ab ≠ 1, c ≠ -d
ab ≠ 1 because otherwise
Example of validity:
log 3 81 = 4, log ⅓ 81 = -4 , log 1 81 = ?
(cd)/(c + d) would be (4X-4) / (4 + -4) = -16/0, which is undefined.
1 raised to any power is always equal to 1, and will never equal 81.
This also shows why c ≠ -d, because you can't divide by 0.
Part 3 of Problem 2:
Create two or more examples that fit the pattern above.
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Log 3 81 log 9 81 log 27 81
log 3 81 = 4, log 9 81 = 2 log27 81 = 4/3
Therefore (cd) / (c + d) = (4X2) / (4 + 2) = 8/6 = 4/3
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log 4 64 log 64 64 log 256 64
log 4 64= 3 log64 64= 1 log256 64 = ¾
Therefore (cd) / (c + d) = (3X1) / (3 +12) = 3/4 = ¾
Generalizations:
- You can find the logarithms of positive and negative integers.
- Logarithms follow exponential rules being exponents themselves.
Limitations:
- Restrictions for certain variables were explained within the problem itself, therefore, view corresponding figures to see those circumstances.
- You can find the log of fractions and square roots as shown in figure 8.
- If trying to find the logarithm of a negative number the result will hold an imaginary solution rather than a “real” number.