Keystone Problems… Keystone Problems… next Set 19 © 2007 Herbert I. Gross.

Keystone Problems…Keystone

Problems…

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Set 19© 2007 Herbert I. Gross

You will soon be assigned problems to test whether you have internalized the material

in Lesson 19 of our algebra course. The Keystone Illustrations below are

prototypes of the problems you’ll be doing. Work out the problems on your own.

Afterwards, study the detailed solutions we’ve provided. In particular, notice that several different ways are presented that could be used to solve each problem.

Instructions for the Keystone Problems

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© 2007 Herbert I. Gross

As a teacher/trainer, it is important for you to understand and be able to respond

in different ways to the different ways individual students learn. The more ways

you are ready to explain a problem, the better the chances are that the students

will come to understand.

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Given that…

100.3010 = 2;

what is the value of x if 104.3010 = x?

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Keystone Problem #1 Keystone Problem #1 for Lesson 19for Lesson 19

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When a number is expressed in the form 104.3010 , the whole number part of the exponent (in this case, 4) tells us the order of magnitude of the number, and the part of the exponent that’s to the right of the whole number (in this case, .3010) tells us the sequence of digits in the number.

Solution for Problem #1Solution for Problem #1

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To see why this is true, we use the fact that 4.3010 = 4 + 0.3010 and rewrite104.3010 in the form…

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By our rule for multiplying like bases, we know that …104 + 0.3010 = 104 × 100.3010. We also know that 104 = 10,000 and

100.3010 = 2. Therefore…

104.3010 = 104 + 0.3010

104.3010 = 104 × 100.3010 = 10,000 × 2 = 20,000

Even without knowing the exact value of 104.3010, the fact that 10x increases as x

increases tells us that since 4 < 4.3010 < 5, it means that

104 < 104.3010 < 105

Notes #1

And since 104 = 10,000 and 105 = 100,000 we see that…

10,000 < 104.3010 < 100,000

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Given that… 100.6990 = 5,

100.4771 = 3 and 100.3010 = 2;

for what value of x will 10x = 30?


One of the nice things about working with exponents is that it allows us to replace multiplication problems by

equivalent addition problems. This was very important in the days before the advent of calculators and

computers.

Preface to the SolutionPreface to the Solution

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For example, using pencil and paper to compute such products as

3.14159 X 2.98764 X 5.12347 would be very tedious to say the least. On the other hand, with a hand calculator we can compute this

product with ease. However, prior to the age of calculators, Napier’s invention of logarithms (arithmetic using exponents)

was a way to replace such tedious multiplication problems by addition

problems.

PrefacePrefacenext


Of course with the advent of calculators

there is no longer a need to use logarithms in order to perform arithmetic calculations. However, it is still very important to know

and to be able to use the rules for exponent arithmetic because so many things in “real

life” involve exponential growth.

PrefacePrefacenext


Using small whole numbers for simplicity, this exercise is designed to illustrate how we can replace a multiplication problem

by an equivalent addition problem.

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We are given the following three pieces of information…

100.6990 = 5

We also know that…

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100.4771 = 3

100.3010 = 2

5 × 3 × 2 = 30

12233

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If we now replace 5, 3, and 2 in the equation 5 × 3 × 2 = 30 by their above values, we see

that…

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5 × 3 × 2 = 30

100.6990100.6990 = 5 100.4771100.4771 = 3 100.3010100.3010 = 2

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By the rule for multiplying like bases, the above equation can be rewritten in the

equivalent form…

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100.6990 × 100.4771 × 100.3010 = 30

100.6990 + 0.4771 + 0.3010 = 30

Since 0.6990 + 0.4771 + 0.3010 = 1.4771, we may rewrite the above equation as…

101.4771 = 30

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Since no two different powers of 10 can have the same output, we see from the

above equation that the only way in which 10x can equal 30 is

if x = 1.4771.

101.4771 = 30

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To check our answer we may use a calculator with an xy key and then use the

following sequence of key strokes…

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Using my calculator the result is101.4771 = 29.99853...

The slight discrepancy between 29.99853… and 30 comes from the fact that the

exponents in the equation 100.6990 × 100.4771 × 100.3010 = 30 have been

rounded off to 4 decimal places.

1 0 xy 1 . 4 7 7 1 =

Information such as 100.6990 = 5 was made available through tables that were

developed by Napier. How he did this is well beyond the scope of this course.

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Notes #2

However, there seem to be a fewinteresting “coincidences” in our solution;

one of which is that the sum of the two exponents 0.6990 and 0.3010 is exactly 1.

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To see why this is not a coincidence, observe that…

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Notes 2

5 × 2 =

100.6990 × 100.3010 =

100.6990 + 0.3010 =

101 =

1010

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Of course we would probably never have

elected to compute2 × 3 × 5 this way. Just imagine not having a calculator and having to compute the product 3.14159 × 2.98764 × 5.1234. However, from the exponential arithmetic (logarithm) tables developed by Napier, we could find that…

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Note #2

100.49715 = 3.14159

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100.47533 = 2.98764

100.70956 = 5.12347

We could then proceed as we did in the previous solution by rewriting…

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Note #2

n = 3.14159 × 2.98764 × 5.12347

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n = 100.49715 + 0.47533 + 0.70956

in the form…

…and by the rules for multiplying like bases we see that…

n = 100.49715 × 100.47533 × 100.70956

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Since 0.49715 + 0.47533 + 0.70956 = 1.68204, we may rewrite our equation as…

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Note #2next

n = 100.49715 + 0.47533 + 0.70956

We may then use the xy key on the calculator to see that n = 48.0883637...

1.68204

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(or in the language of logarithms, n = log 1.68204).

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AsideAt the time of Napier, there were no calculators. His tables were designed to find the logarithm of a given number.

Thus, if a person had been presented with the problem log n = 1.68204, he or she would have had to make an indirect computation. That is: starting with the logarithm of a number (1.68204), they would then have to work “backwards”

in the table in order to find the number.

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Note #2

If we were to use “pencil and paper” to find the exact value of the product

3.14159 × 2.98764 × 5.12347, we would have found that its exact value was

48.088581743330172.

The discrepancy between the exact value and the value we got by using the log table,

is that the numbers in the table were rounded off. As a result, only the first few

decimal places in our estimated answer were significant.

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Note #248.0883637 48.088581743330172

Table Exact

The above discussion serves to illustrate why the study of significant figures was

so important before the advent of calculators.

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There is much more information that we can derive from the fact that

100.3010 = 2

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Note #2

10x = 8

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For example, suppose we wanted to find the value of x for which…

We know that 8 = 23. Therefore, we may rewrite the above equation as…

10x = (2)3

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We also know that 100.3010 = 2. Therefore, we may rewrite the equation

10x = (2)3 as…

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Note #2

10x = (100.3010)3

By our rules for the arithmetic of exponents we know that…

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(100.3010)3 =

10(0.3010)3 =

100.9030

Thus, replacing 8 by 100.9030, the equation 10x = 8 becomes…

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Note #2

10x = 100.9030

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Since no two different powers of 10 can be equal, we see from the above

equation that x = 0.9030.

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As a check, we may use our calculator (or tables) to see that 100.9030 = 7.99... which

tells us that our answer is at leastplausible and probably correct.

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Given that…

100.3010 = 2 and 101.3802 = 24;

what is the value of x for which 2x = 24?

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Most calculators come equipped with a 10x key (or, at least an xy key).

However, they do not have other keys such as 2x or 3x (let alone a key such as 3.7x etc.), but with the use of the 10x key there is no need for having these other

keys.

Preface to the SolutionPreface to the Solutionnext


With the rules for exponents all we would need is any one key of the form bx.

However, b was chosen to be 10 primarily to take advantage of place value.

PrefacePrefacenext


For example, when we see 105.1567 we know immediately that the number is

between 105 and 106; that is: it is between 100,000 and 1,000,000.

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With the 10x key, we can let x = 0.3010 and verify that 100.3010 = 2. However,

other than for trial-and-error methods the 10x key can’t help us find the value

of x if we know the value of 10x.

PrefacePrefacenext


For example, suppose we were given a problem such as:

Find the value of x for which 10x = 7.

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In fact, to find the value of x for which 10x = 7, we would want to use the

“undoing” method. (That is: in the same way that we use subtraction to undo addition we would like to find a

way to “undo” 10x.)

PrefacePrefacenext


In this regard, calculators that have a 10x key also have another key that is

labeled log.

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Let’s look at a rather simple illustration in terms of the equality 103 = 1,000. Namely, if we are told that 103 = x, to

find the value of x using the calculator we would enter “3” and then press the

10x key to obtain x = 1,000.

PrefacePrefacenext


On the other hand, if we were told that 10x = 1,000, we would enter “1,000” and then press the log key to obtain x = 3.

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It is not important for you to internalize, the full meaning of “logarithm”;

just remember that it is simply an abbreviation for a specific exponent.

LogarithmsLogarithmsnext


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For example, an equality such as log 56 = 1.748 is simply another way of

saying that 101.748 = 56

That is: “log x” is an abbreviation for “the power to which 10 has to raised to

yield x as the answer”.

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More generally… An equivalent way of saying that 10x = y

is to say that log y = x.



In other words, if you understand the meaning of say, 104 = 10,000, but you don’t understand the meaning of log 10,000 = 4;

it means that you are having a problem with the new vocabulary; not with mathematics.

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LogarithmsLogarithms

Applying this idea to solving for x if 10x = 7, we simply enter 7, press the log key and then the “=” key. That is, we enter…

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…and 0.8450983… appears in the display. As a check, using the 10x key and rounding

off 0.8450983… to 0.845098, we see that100.845098 = 6.9999993...; which validates that

our answer is quite reasonable.

7 log =

Exercise 3 illustrates how by using the 10x and log keys, we can solve the problem

2x = 24,



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We are told that 100.3010 = 2 and 101.3802 = 24

Hence, we may replace 2 by 100.3010 and 24 by 101.3802 in the equation 2x = 24 to obtain the equivalent equation…

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(100.3010)x = 101.3802.

We know from our rules for exponents that (100.3010)x = 101.3010x.

100.3010x = 101.3802

Hence, we may rewrite the above equation as…

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Since no two different powers of x yield the same value for 10x, we may conclude from the above equation that 0.3010x = 1.3802;

or x = 1.3802 ÷ 0.3010; and using the calculator for convenience we see that…

x = 4.58539…

100.3010x = 101.3802

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To check that our solution is correct we may round off 1,3802 ÷ 0.3010 to 4.58539 and replace x by 4.58539. Doing this we

see that 24.58539 = 24.00711…; which indicates that the solution of the equation

2x = 24 is approximately x = 4.58539.

Ordinarily in trying to solve the equation 2x = 24, we would not have had the luxury

of knowing in advance that: 100.3010 = 2 and 101.3802 = 24. However, we could have obtained this information by

using the log key to see that log 2 = 0.301029995… and

log 24 = 1.38021134… (which is even more accurate than the given information)

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Note #3

So let’s go through the procedure one more time, and this time find the value of

x for which… 11x = 212

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Note #3

We use the log key to see that… log 11 = 1.04139… or 101.04139 = 11

and log 212 = 2.32633… or 102.32633 = 212

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So if we now replace 11 by 101.04139 and 212 by 102.32633, the equation 11x = 212 can be

replaced by the equivalent equation…

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Note #3

…and we may therefore conclude that 1.04139x = 2.32633

(101.04139)x = 102.32633

Hence, x = 2.32633 ÷ 1.04139 = 2.23387…

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As a check that the solution of 11x = 212 is x = 2.23387…, we approximate 2.23387… by 2.23387 and then replace x by 2.23387

in the equation 11x = 212. We then see that

112.224267 = 212.0000081… and this indicates that we have found the correct answer, at least to several decimal place accuracy.

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Note #3

In solving the equation 11x = 212, we found that x = 2.32633 ÷ 1.04139 = 2.23387.

Let’s look at this quotient in more detail.

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Namely, 2.32633 = log 212 and 1.04139 = log 11.

Hence, what we have shown is that…

If 11x = 212, then x = log 212 ÷ log 11.

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Enrichment Note

More generally, the log key allows us to solve any equation of the form bx = n

where b and n are any positive numbers.

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Namely if bx = n, then x = log n ÷ log b

For example, in terms of 2x = 24, we saw that x = 1.3802 ÷ 0.3010, and since log 24 = 1.3802 and

log 2 = 0.3010, we see thatx = log 24 ÷ log 2

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Enrichment Note

Keystone Problems… Keystone Problems… next Set 19 © 2007 Herbert I. Gross.

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Transcript of Keystone Problems… Keystone Problems… next Set 19 © 2007 Herbert I. Gross.