## SAT SUBJECT TEST MATH LEVEL 1

## TOPICS IN ARITHMETIC

**CHAPTER 2**

**Basic Arithmetic**

**LOGARITHMS**

Recall that the statement 2^{4} = 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 2

^{4}= 16 is equivalent to the statement log

_{2}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,**

**log**_{b}** x** =

*y***if and only if**

*b***=**

^{y }

*x*.For example:

• log_{10}100 = 2 because 10^{2} = 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, log_{10} 5 is the number *x* such that 10* ^{x}*= 5. Since 10

^{0}= 1 and 10

^{1}= 10, you know that 0 < log

_{10}5 < 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 log

_{10}5 = LOG 5 = 0.699.

What if you want to know the value of log_{2} 5? Clearly, you cannot just enter LOG 5 on your calculator since that means log_{10} 5. Also, since 2^{2} = 4, log_{2} 4 = 2 and so log_{2} 5 > 2, certainly not 0.699. The following KEY FACT shows you how to use your calculator to evaluate log_{2} 5 exactly.

**TIP**

The LOG button on your calculator evaluates *only* base 10 logarithms. So LOG 5 = 0.699 means log_{10} 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 log_{10}5 and log_{10}3, 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*:• **log**_{b}**( xy)** =

**log**

_{b }**+**

*x***log**

_{b }

*y*•

• **log**_{b}*x***^{n}**=

*n***log**

_{b }

*x*• **log**_{b }*b***^{n}**=

*n***(In particular, log**

_{b }**1**=

**log**

_{b }

*b***=**

^{0 }**0 and log**

_{b }**=**

*b***log**

_{b }

*b***=**

^{1}**1)**

For example, if log* _{b }*2 =

*x*and log

*3 =*

_{b }*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 2* ^{x }*= 512, for example, take the logarithm of both sides: log

*2*

_{b }*= log*

^{x }*512. Then*

_{b }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