• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

Logarithm Bases

Extracts from this document...

Introduction In this analysis of logarithm bases I will be discussing and explaining some of the properties and functions dependent on formulas which can be obtained by the use of logarithms. Simply, a logarithm is just an exponent. View the following diagram below in figure 1. An alternate explanation is that, the logarithm of a number x to a base b is just the exponent you put onto b to make the result equal x. For instance, since 5² = 25, we know that 2 (the exponent to the five) is the logarithm of 25 to base 5. Represented as, log5(25) = 2.

Figure 1.

In a general expression, when applying variables to replace numbers, if x = by, then we say that y is “the logarithm of x to the base b” or “the base b logarithm of x”. Represented as,

y = logb(x). Therefore, any exponential equation can be written as a log just by changing the x and y positions to manipulate the equation.

Another way to look at it is that the logbx function is defined as the inverse of the bx function. These two statements express that inverse relationship, showing how an exponential equation is equivalent to a logarithmic equation. View the diagrams in figure 2 and figure 3.  Figure 2.                                                        Figure 3.

Middle

"The logarithm of a power of x is equal to the exponent of that power
times the logarithm of x.
"
1. logbxn  =  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.

 base oflogarithms symbol name 10 log(if no base shown) common logarithm e ln natural logarithm,pronounced “ell-enn” or “lahn” Figure 15.

The “base”ics

From the

Conclusion

-1/1                        log 1/625 125    = -3/4  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  81 = -4 , log 1 81 = ?
(cd)/(c + d) would be (4
X-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.

• 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) = (4
X2) / (4 + 2) = 8/6 = 4/3
• log 4 64                 log 64 64           log 256 64
log
4 64= 3            log64 64= 1       log256 64 = ¾

Therefore (cd) / (c + d) = (3
X1) / (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.

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

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

Related International Baccalaureate Maths essays

1. Mathematics (EE): Alhazen's Problem

The boundary in our case would be a tangent line drawn to the point on the border of the circle where the ball A bounces off the circular side to ball B (Figure 2). Another way to express this problem is, "to describe in a given circle an isosceles triangle

2. Logarithms. In this investigation, the use of the properties of ...

log21(23) = log23 / log21 log22(23) = log23 / log22 log24(23) = log23 / log24 log25(23) = log23 / log25 Then, according to the power property of logarithms, the exponent of a logarithm can be moved to the front of the log as a multiplier which can be further simplified by canceling out log2 on each side of the fraction.

1. In this investigation I will be examining logarithms and their bases. The purpose of ...

Thus, the expressions above can also be proven using a graph. Figure 1.1 shows 4 graphs and their points of intersection. Y= s(x), where s(x) = Y= t(x), where t(x) = log28 Y = u(x), where u(x) = log168 Y = v(x), where v(x)

2. matrix power

Sometimes the zeros could be replaced by a different number. In this case the power of the non-zero digit, being 2 in this report, replaces the non-zero digit in the identity matrix, and the zeros remain the same. Therefore the general expression of a Mn matrix in terms of "n

1. Math IA Logarithm bases

N being the nth term of the sequence. 1st Log 3 81 = 4 4/1 = 1 2nd Log 9 81 = 2 4/2 = 2 3rd Log 27 81 = 1.33 4/3 = 1.33 4th Log 81 81 = 1 4/4 = 1 5th Log 243 81 = .8 4/5 = .8 6th Log 729 81 =

2. This essay will examine theoretical and experimental probability in relation to the Korean card ...

Experimental Probability The number of trial is the total number of times the experiment is repeated. The outcomes are the different results possible for one trial of the experiment The frequency of a particular outcome is the number of times that this outcome is observed.

1. Arithmetic Sequence Techniques

I. 2 (7) and (5 + 9); 2(11) and (9 + 13); 2 (13) and (11 + 15) II. 2 (7 + 9) and (3 + 5) + (11 + 13) III. 2 (11 + 13 + 15) and (5 + 7 + 9)

2. IB Pre-Calculus Logarithm Bases General Information: Logarithms A ...

In the first sequence, the next three terms are: log64 8, log128 8, and log2568. In the second sequence the next four terms are: log243 81, log729 81, log2187 81, and log6561 81. In the third sequence, the next four terms are: log3125 25, log15625 25, log78125 25, and log390625 25. • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to
improve your own work 