﻿ ﻿LOGARITHMS - Basic Arithmetic - TOPICS IN ARITHMETIC - SAT SUBJECT TEST MATH LEVEL 1

## TOPICS IN ARITHMETIC ## CHAPTER 2Basic Arithmetic ### LOGARITHMS

Recall that the statement 24 = 16 is usually read as “2 to the 4th power equals 16.” Another way to read this stresses the role of the exponent 4: “4 is the exponent to which the base 2 must be raised to equal 16.” Mathematicians have a special word for this exponent—logarithm. The statement 24 = 16 is equivalent to the statement log2 16 = 4, which is read, “the base 2 logarithm of 16 is 4.”

TIP A logarithm is an exponent.

Key Fact A16

If b is a positive number not equal to 1 and x > 0,

logbx = y if and only if b y = x.

For example:

• log10100 = 2 because 102 = 100 because  because Although you can always estimate the value of a logarithm, in general there is no easy way to evaluate a logarithm exactly without a calculator. For example, log10 5 is the number x such that 10x= 5. Since 100 = 1 and 101 = 10, you know that 0 < log105 < 1. You could improve your estimate by noting that since , which is only slightly greater than 3, , and so 0.5 < x < 1. The LOG button on your calculator evaluates base 10 logarithms. So you can use your calculator to get that log10 5 = LOG 5 = 0.699.

What if you want to know the value of log2 5? Clearly, you cannot just enter LOG 5 on your calculator since that means log10 5. Also, since 22 = 4, log2 4 = 2 and so log2 5 > 2, certainly not 0.699. The following KEY FACT shows you how to use your calculator to evaluate log2 5 exactly.

TIP The LOG button on your calculator evaluates only base 10 logarithms. So LOG 5 = 0.699 means log10 5 = 0.699, which is often written log 5, without the base being indicated.

Key Fact A17

CHANGE OF BASE FORMULA  Similarly, and .

Reminder: In the line above, log 5 and log 3 are abbreviations for log105 and log103, respectively.

You need to know a few important laws of logarithms. They are all direct consequences of the laws of exponents (KEY FACT A11).

Key Fact A18

LAWS OF LOGARITHMS

For any positive base b 1 and any positive numbers x, y, and n:

logb(xy) = logb x + logb y logbxn= n logb x

logb bn= n (In particular, logb 1 = logb b0 = 0 and logb b = logb b1= 1)

For example, if logb 2 = x and logb 3 = y, then: As you will see in Chapter 6, the third rule in KEY FACT A17 allows you to solve equations in which the variable is an exponent by bringing the variable down to the base line. To solve the equation 2x = 512, for example, take the logarithm of both sides: logb 2x = logb 512. Then Note that what you choose for the base b does not matter. As a consequence, you might as well let b = 10. Therefore, . Now use your calculator:

x = (log 512) (log 2) = 9

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