## Learning Math: Number and Operations

# Place Value Part B: Exponents and Logarithms (35 minutes)

As you saw earlier in this session, one way to represent each power of two is to write the base (two) raised to a power (another number). This other number is known as the exponent. An exponent tells us how many times the base is used as a factor. Exponents can simplify the calculations for such operations as multiplication and division. For example, rather than multiply 16 • 32, we can multiply 2^{4} • 2^{5}. Let’s look at how this is done.

To compute with numbers that have exponents, you need to understand how exponents work. Here are some basic rules to begin with:

- To add or subtract numbers with exponents, the base numbers must be the same, and the exponents must also be the same:

**x**^{4}+ x^{4}+ x^{3}+ x^{3}= 2x^{4}+ 2x^{3} - To multiply numbers with exponents, the base numbers must be the same; then we simply add the exponents. For example:

**x**^{4}• x^{3 }x^{4}= x • x • x • x and x^{3}= x • x • x

Because multiplication is both associative and commutative, we can solve these equations as one:

**x**^{4}• x^{3}= x • x • x • x • x • x • x = x^{7 }So the end result, x^{7}, is equivalent to x^{4 + 3}:

**x**This presumes that both the bases are the same. In other words, for example, we couldn’t multiply 2^{4}• x^{3}= (x • x • x • x) • (x • x • x) = x^{4 + 3}= x^{7 }^{2}by 3^{3}because the bases are not the same.

- To divide numbers with exponents, the base numbers must be the same; then we simply subtract the exponents. For example:

These examples illustrate the meaning of positive-integer exponents. But what does an exponent of 0 represent?

**Problem B1**

Use the rule x^{a} x^{b} = x^{a-b} to figure out the value of x when the exponent is 0.

If a – b = 0, what does that tell you about a and b?

**Problem B2**

What happens if the exponent is a negative integer like -1? Solve x^{3}÷x^{4} to find out. Explain why x cannot be equal to 0.

Now let’s look at what happens when the exponent is a fraction or a decimal. We know that for positive numbers greater than or equal to 1, x^{2} x^{3} x^{4}. Is this true for exponents between 0 and 1?

**Problem B3**

**a. **Use the rules for multiplying exponents to determine the meaning of x^{1/2}.

**b. **How about x^{1/3}?

**c.** Which value is greater for positive numbers greater than 1?

**Problem B4
**Express (x

^{3})

^{2}as a multiplication problem and then simplify it as much as possible.

**Problem B5**

Consider your answers to Problems B1-B4. In each case, can 0 be a valid base? Explain why (or why not).

**In This Part: Scientific Notation
**When doing complicated computations with very large or very small numbers, people often write the numbers in scientific notation so that they can more easily estimate the magnitude of the number. In scientific notation, every number is written as a decimal number greater than or equal to 1 but less than 10, multiplied by a power of 10. Thus, the decimal always has exactly one non-zero digit to the left of the decimal point. For example, 6,253 would be written 6.253 • 10

^{3}, and .06253 would be written 6.253 • 10

^{-2}. The best way to remember how to find the correct power of 10 is to write the number with one digit to the left of the decimal point and then think about what to do to that decimal to make it equal to the original number.

**Problem** **B6
**Write the following numbers using scientific notation:

**a.**43,007

**b.**0.00245

**c.**-675

Scientific notation can help us quickly perform operations on large numbers. Calculating 2,300,000 • 3,000,000,000 is much easier to think of as the following:

(2.3 • 10^{6}) • (3.0 • 10^{9}) = 6.9 • 10^{15
}

**Problem B7**

Perform the following calculations using scientific notation:

**a. **2,300,000 + 790,000

**b. **10,000,000 • 678,000,000,000

**c. **1,490,000,000 7,000

**In This Part: Logarithms
**John Napier from Scotland invented logarithms in the early 17th century. Napier was not a professional mathematician, but he made many important contributions to mathematics. He invented not only logarithms, but the decimal point as well, and he carved multiplication tables on sticks to simplify the multiplication of multi-digit numbers.

What are logarithms? Basically, they are exponents. In order to use logarithms, you must stipulate both the base and the exponent. Here is one example: Since 2^{3} is equal to 8, we can write in symbols log_{2} 8 = 3, which we read as “log to the base two of 8 is 3.” This means that the exponent needed on the base two to get to 8 is 3. When working in base ten, it is not necessary to write the base. For example, log 50 is the same as log_{10} 50.

Before the advent of calculators and computers, logarithms were extremely important because they simplified complex multiplication and division by turning them into simple addition or subtraction, and reduced powers to multiplication. See Note 1 below.

Most calculators are programmed with values for base ten (abbreviated LOG) and base *e*(abbreviated LN) logarithms. The letter *e* represents the transcendental number that is the base of natural logarithms. The value of *e* is found by taking the limit of (1 + 1/n)^{n}as n approaches infinity. This gives the value of about 2.718.

**Problem B8**

**a.** What is log 100?

This question is asking, what exponent is needed in base ten to get 100? In other words, if 10^{x} = 100, what is x?

**b. **What is log_{3} 81?

**c. **What is the base ten number whose log is 4?_{
}**d. **What is log_{b} b?

**Problem B9
**

**a.**Estimate the value of log

_{5}50.

**b.**Estimate the value of log

_{3}100.

These logarithms will be decimals. Try to figure out between which integers these fractions will fall.

### Solutions

**Problem B1**

One way to do this division is to use the rule that x^{a} x^{b} = x^{a – b}. Here, this means that x^{3} divided by x^{3} equals x^{0}. The other way to do this is to recognize that we are dividing a number by itself and that any non-zero number divided by itself equals 1:

Since the answers must be the same, this means that x^{0} = 1 for any value of x except 0.

**Problem B2
**One way to do this division is to use the rule that x

^{a}x

^{b}= x

^{a – b}. Here, this means that x

^{3}divided by x

^{4}equals x

^{-1}. The other way to do this is to write out the numerator and denominator:

If x is not 0, we can cancel x three times from the numerator and denominator to leave 1/x as the final answer. Since the answers must be the same, this means that x^{-1} = 1/x for any value of x except 0 (1/x is undefined for x = 0). Therefore, in any division problem involving a negative exponent, we must restrict the base to a non-zero number.

**Problem B3**

**a. **If x^{1/2} follows the same rules as x^{m} for integer m, then x^{1/2} multiplied by x^{1/2} must be x. This means that x^{1/2 }is the number we multiply by itself to make x. This is the definition of a square root, so x^{1/2} represents the square root of x.

**b. **Similarly, x^{1/3} needs to be multiplied by itself three times to make x, so it is the cube root of x.

**c.**If x is a positive number greater than 1, then x^{1/2} will be greater than x^{1/3}. One way to think about this is to look at x^{1/6}, the sixth root of x. Since x is greater than 1, x^{1/6} is also greater than 1. If x^{1/6} were positive but less than 1, multiplying it by itself would give a smaller number. Since x^{1/6} is greater than 1, multiplying it by itself produces a larger number each time.

If you multiply x^{1/6} by itself, or x^{1/6}, you get x^{2/6}, or x^{1/3}. Meanwhile, x^{1/6}multiplied by itself three times produces x^{3/6}, or x^{1/2}. Therefore, x^{1/2} must be larger than x^{1/3} if x is a positive number greater than 1:

Therefore, we know that x^{0} < x^{1/6} < x^{1/3} < x^{1/2} < x^{2/3} < x^{1}.

(The points on the line above are not drawn to scale.)

**Problem B4
**To do this, we write out exponentiation as repeated multiplication: (x

^{3})

^{2}= x

^{3}• x

^{3}= (x • x • x) • (x • x • x). According to the rules given earlier, we add these exponents when multiplying, so the result is x

^{6}, the same value as x

^{3 • 2}.

In general, consider(x^{a})^{b}. Writing this as a multiplication problem, we’d see x^{a} • x^{a} • x^{a}… • x^{a}, where there are b occurrences of x^{a}. Adding these exponents, we get a + a + a … + a; that is, a added to itself a total of b times. Repeated addition is multiplication, so the result is ab, the product of a and b. So we see that, in general, (x^{a})^{b} = x^{(a • b)} = x^{ab}

**Problem B5
**Zero cannot be used as a base for several reasons. The base can only be 0 when working with rules involving multiplication and exponentiation of positive exponents. However, all positive powers of 0 equal 0, and products and sums of 0 are all 0, thus making a one-value system. Since we cannot divide by 0, we cannot define 0

^{0}as 0

^{n}0

^{n}for some n (see Problem B1 for more information). Additionally, we cannot define 0

^{n}for any negative n (see Problem B2).

**Problem B6**

**a. **43,007 = 4.3007 • 10^{4}. The lead digit (4) was originally in the 10,000, or 10^{4}position, so when we move it to the 10^{0} position, we must multiply by 10^{4}. Multiplying 4.3007 by 10^{4} gives us 43,007.

**b. **0.00245 = 2.45 • 10^{-3}. The lead digit (2) was originally in the 1/1000, or 10^{-3}position, so when we move it to the 10^{0} position, we must multiply by 10^{-3}. Multiplying 2.45 by 10^{-3} gives us 0.00245.

**c. **-675 = -6.75 • 10^{2}. The lead digit (6) was originally in the 100, or 10^{2} position, so when we move it to the 10^{0} position, we must multiply by 10^{2}. Multiplying -6.75 by 10^{2} gives us -6.75.

**Problem B7
**

**a.**2,300,000 + 790,000 = (2.30 • 10

^{6}) + (.79 • 10

^{6}) = 3.09 • 10

^{6 }

**b.**10,000,000 • 678,000,000,000 = (1 • 10

^{7}) • (6.78 • 10

^{11}) = 6.78 • 10

^{7+ 11}= 6.78 • 10

^{18 }

**c.**1,490,000,000 7,000 = (1.49 • 10

^{9}) (7 • 10

^{3}) = (1.49 7) • 10

^{9 – 3}= 0.213 • 10

^{6}

**Problem B8**

**a. **In exponential form, log 100 is 10^{x} = 100. Since 10^{2} = 100, log 100 = 2.

**b. **In exponential form, log_{3} 81 is 3^{x} = 81. Since 3 • 3 • 3 • 3 = 3^{4} = 81, the solution is 4.

**c. **If log x = 4, the equation in exponential form is 10^{4} = x, so x = 10,000.

**d. **The equation is b^{x} = b. This is solved by x = 1 for all valid bases for b (with the convention that b must be positive and not equal to 1).

**Problem B9
a. **Log

_{5}50 = x. We can also write this expression as 5

^{x}= 50, i.e., 5 raised to what power equals 50? We know that 5

^{2}= 25 and 5

^{3}= 125, so we can estimate x as somewhere between 2 and 3. To bring our estimate even closer, we can see what happens for x = 2.5, which, converting it into a fraction, can be written as 25/10, or 5/2. So we write 5

^{5/2}= = 55.9. Thus, we can further narrow down our estimate by saying that x is slightly under 2.5.

**Similarly, log**

b.

b.

_{3}100 = y can be written as 3

^{y}= 100. We know that 3

^{4}= 81 and 3

^{5}= 243, so we can estimate y as somewhere between 4 and 5. For y = 4.5 we can convert 4.5 into a fraction by writing 4.5 = 45/10 = 9/2, we get 3

^{9/2}= = 140.3. Thus, we can further narrow down our estimate by saying that y is somewhere between 4 and 4.5.