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# Logarithm Bases

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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.

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