• The inverse of the y = e^x function: table and graph                                    Exercises 1-4
• The inverse of the y = b^x function: table and graph                                   Exercises 5-6
• Base-10 logs                                                                                                        Exercises 7-8
• The rules of logarithms and solving logarithmic equations                        Exercises 9-12

A table for y = e^x, for x = -3 to 3, is shown below:

x	y = e ^ x
-3	.0498
-2	.1353
-1	.3679
0	1.000
1	2.718
2	7.398
3	20.09

Reversing this table we get a partial table for the inverse function y = log{base e} (x):

x	y = log{base e} (x)
.0498	-3
.1353	-2
.3679	-1
1.000	0
2.718	1
7.389	2
20.09	3

Graphing the base-e exponential function y = e^x and the corresponding logarithmic function y = log{base e} (x) on the same set of axes as the base-2 exponential and logarithmic functions y = 2^x and y = log{base 2} (x), we obtain the following picture:

To see how the two exponential graphs compare, consider the basic points (0,1), common to both graphs, and the points (1,2) on the y = 2^x graph and (1,2.718...) on the y = e^x graph. It is clear that the base-e function increases more quickly for positive x values than the base-2 function, while approaching the common asymptote x = 0 more quickly for negative x values.

The logarithmic graphs are as usual obtained by reversing the coordinates of the exponential graphs, which has the effect of reflecting the exponential graphs across the y = x line. Note that the x and y scales on the above graph are not equal, so that the y = x line doesn't make a 45-degree angle and the line segments connecting corresponding points between an exponential and logarithmic graph will not appear to cross the y = x line at right angles.

The reversed log-function versions of the basic points (1,2) and (1, 2.718...) of the exponential graphs are shown on the appropriate logarithmic graphs. We note that the logarithmic graphs both pass through the point (1,0), which is the reflection of the common point (0,1) of the exponential functions, and they both reach the value y = 1 when x is equal to the base (i.e., 2 or e = 2.718...) of the logarithmic function.

Just as the graphs of the exponential functions cannot pass below the negative x axis, the graphs of the logarithmic functions cannot pass through the negative y axis.

The function y = log{base e} (x) is called the natural log of x, and is written y = ln(x). The 'ln' stands for 'log natural'. Some people pronounce 'ln' as 'line', which doesn't seem to be a good idea for two reasons: many mathematicians are unfamiliar with this usage and won't know what you are talking about, and 'line' has a pretty specific mathematical usage that has nothing to do with logarithms of any sort. However, if you encounter people using this pronunciation (and many fine people, including some excellent mathematicians and teachers, do use it), it is usually best just to humor them (recall that the unabomber is also a mathematician of extraordinary skills).

The following statement emphasizes the 'ln' notation:

This convention appears on the keyboards of most calculators, with the y = e^x function and the y = ln(x) function sharing the same key. One function is obtained by pressing the key, the other by pressing the inverse key then the key.

1.   For what value of x will the function y = log{base 2}(x) first reach 4 (hint: what do you know about the coordinates of the corresponding point on the graph of y = 2^x?). For what value of x will the function y = ln(x) first reach y = 4? For what x values will it first reach y=2 and y=3?

2.  One of the graphs contains an obvious error. Which graph is it and what is the error? (Hint: look at asymptotes).

3.  Explain why the negative y axis is an asymptote for a log{base b}(x) function (hint: what do you get when you reverse the coordinates of a point on the negative x axis?).

4.  Sketch a graph of y = ln(x) for 0 < x <= 100. Use your graph to estimate the natural logs of 10, 20, 30, ..., 100.

If b is any positive number, the exponential function y = b^x will have a corresponding inverse y = log{base b} (x).

A table for y = b^x might be as follows:

x	y = b^x
-3	b^-3
-2	b^-2
-1	b^-1
0	b^0 = 1
1	b^1 = b
2	b^2
3	b^3

The corresponding logarithmic function would have the following partial table:

x	y=log{base b} (x)
b^-3	-3
b^-2	-2
b^-1	-1
1	0
b	1
b^2	2
b^3	3

Typical graphs are shown below:

5.  Estimate the values of b for the two exponential functions on the above graph. At what points will each of the logarithmic functions reach the values y = 2, y = 3 and y = 4?

6.  Sketch an approximate graph of y = log{base 5} (x), for 0 < x <= 100.

The inverse function of y = 10^x is y = log{base 10} (x). Base-10 logs are very basic to our numbering system, which is based on 10. In this sense they seem more natural to us than what we call natural logs (base-e logs). However base-10 logarithms lack the convenient mathematical properties of base-e logs, so base e deservededly gets the label 'natural'.

Partial tables for base-10 exponential and logarithmic functions are shown below:

x	y = 10^x
-3	.001
-2	.01
-1	.1
0	1
1	10
2	100
3	1000

 

x	y=log{base 10} (x)
.001	-3
.01	-2
.1	-1
1	0
10	1
100	2
1000	3

Base-10 logs can be written in the form log{base 10} (x), but we usually just drop the {base 10} and write log(x). Whenever the expression log(x) appears without a specified base, the base is understood to be 10. log(x) means log{base 10} (x).

It is often convenient to represent rapidly growing quantities, or quantities that involve a great range of large numbers, by their logarithms. In fact, our senses seem to experience the world on a logarithmic scale.

For example, the loudest sound a young person with unimpaired hearing can stand without pain is 1,000,000,000,000 (that's a trillion, a million millions) times as loud as the quietest detectable sound. So how loud does a sound seem which is 1,000,000 times as loud as that quietest sound? This sound is a million times louder then the quietest sound detectable, but a million times quieter than the loudest sound we can stand. It is therefore only a millionth of the way from the quietest to the loudest. So does it seem to us that it is closer to the quietest or to the loudest sound?

It turns out that this sound seems to us to be halfway from the quietest to the loudest. It is the ratios of the sounds that we end up comparing, and the ratios are both a million to one.

So let's look at the logarithms of these sound intensities. The three sounds we are interested in are the quietest sound, the sound which is 1,000,000 times as loud as the quietest sound, and the sound which is 1,000,000,000,000 times as loud as the quietest sound. The logarithms are:

1 times quietest sound: log(1) = 0

1,000,000 times quietest sound: log(1,000,000) = 6

1,000,000,000,000 times quietest sound: log(1,000,000,000,000) = 12

The logarithms 0, 6 and 12 are uniformly spaced. So it is the logarithms that correctly model our perceptions.

We often use a 'decibel' scale for models of perception and other quantities that behave in similar weights. The decibel scale is defined as:

decibel level = dB = 10 * log( intensity / threshold intensity),

or

dB = 10 log( I / Io), where Io stands for 'intensity 0', or threshold intensity.

7.  What is the decibel level of a sound which is 10,000 times as loud as hearing threshold intensity (the hearing threshold is the intensity of the quietest sound we can perceive)?

What are the decibel levels of sounds which are 100, 10,000,000 and 1,000,000,000 times louder than threshold intensity?

What are the decibel levels of sounds which are 500, 30,000,000 and 7,000,000,000 times louder than threshold intensity?

8.  If a sound measures 40 decibels, then what is the intensity of the sound, as a multiple of the hearing threshold intensity?

Answer the same question for sounds measuring 20, 50, 80 and 100 decibels.

Write the equation you would solve to find the intensity for each of the following decibel levels: 35, 83 and 117 dB.

The rules of logarithms are directly related to the rules of exponents. The most fundamental of the rules states that

log{base b} (b^x) = x,

which simply says that the base-b logarithmic function is inverse to the base-b exponential function.

The remaining rules are based on this rule and on the rules of exponents. Each rule of exponents has its expression in terms of logarithms.

The rules of logarithms are given below. The rules identified as 'primary' are in fact all the rules that are required. The 'secondary' rules are just convenient versions of the 'primary' rules.

Rule	Corresponding  rule for exponents	Main thought/question
log{base b} (b^x) = x		log is inverse function by definition
log(ab) = log(a) + log(b)	x^a * x^b = x^(a+b)	what does addition have to do with multiplicatioon?
log(a^b) = b log(a)	(x^a)^b = x^(ab)	what does multiplication have to do with exponentiation?

Rule	Idea
log(a/b) = log(a)-log(b)	log(a/b) = log(a*b^-1) = log(a) + log(b^-1) = log(a) + (-1) log(b) = log(a) - log(b)
b ^ [log{base b} (b) ] = b	b^2 and log{base b} (x) are inverse functions
log{base b} (x) = log(x) / log(b)
log(1) = 0	b^0 = 1 for any b

Examples of application of rules:

Example 1: The expression log(x * y * z) can be simplified as follows:

log(x * y * z)

= log( x* (y*z) ) since the multiplication can be grouped in any way we choose

= log(x) + log(y*z) by the law for log(ab)

= log(x) + log(y) + log(z) by the law for log(ab)

So we see that the log of any chain of multiplication can be converted to the sum of the separate logs.

Example 2: An expression like a log(x) + b log(x) becomes log(x^a) + log(x^b), which in turn becomes log(x^a * x^b), which is equal to log(x^(a+b)).

Alternately we could have said a log(x) + b log(x) = (a + b) log(x) = log(x^(a+b)).

Example 3: To find log{base 3} (12), we first write the expression as log(12) / log(3). We then use a calculator or computer to find log(12) and log(3), and calculate our result.

Example 4: To find log{base 3} (9 * 27), we could note that 9 and 27 are powers of 3. If we write

log{base 3} (9 * 27)

= log{base 3} (9) + log{base 3} (27)

= log{base 3} (3^2) + log{base 3} (3^3)

we can obtain our result directly, since log{base 3} (3^2) = 2 and log{base 3} (3^3) = 3.

Example 5: We can find new logarithms from the values of known logarithms. For example suppose that we know that log(16) = 1.204, and log(27) = 1.431. We could then determine the following:

log(16 * 27) = log(432)

= log(16) + log(27) = 1.204 + 1.431 = 2.635,

so log(432) = 2.635.

Since log(27) = log(3^3) = 3 log(3), we have

3 log(3) = 1.431, so

log(3) = 1.431 / 3 = .477.

We could then find log(9) = log(3^2) = 2 log(3), then from this log(81) = log(9^2) = 2 log(9), etc.

We could similarly use the fact that 16 = 4^2 to find log(4), or the fact that 16 = 2^4 to find log(2). These values could be combined with other known values to obtain the logs of many numbers.

Example 6: We can use the properties of logarithms to solve equations in which the desired variable occurs in an exponent.

Consider for example the equation 2 ^ (-3x) = 12.

If we take the logarithm of both sides the exponent -3x will become simply a factor, and we can use simple algebra to solve for x:

2 ^ (-3x) = 12

log( 2 ^ (-3x) ) = log(12)

-3x log(2) = log(12)

x = log(12) / ( -3 log(2) ).

Example 7: We can solve the equation 3 ^ (2x-5) = 2 ^ ( 5x+1) by taking the log of both sides:

3 ^ (2x - 5) = 2 ^ (5x + 1)

log [ 3 ^ (2x - 5) ] = log [ 2 ^ (5x + 1) ]

(2x - 5) log(3) = (5x+1) log(2)

2x log(3) - 5 log(3) = 5x log(2) + log(2).

At this point we stop doing what comes naturally and think for a minute. We finally realize that the x terms need to all be on one side of the equation, and we rearrange to obtain

2x log(3) - 5x log(2) = log(2) + 5 log(3).

This still looks sort of intimidating. But what we have to realize is that all the logs are simple numbers we can find on our calculators. So with some confidence we proceed to factor x out of the terms of the left-hand side:

x ( 2 log(3) - 5 log(2) ) = log(2) + 5 log(x).

Again attempting to hang onto our wits, we divide both sides by (2 log(3) - 5 log(2) ) in an attempt to isolate x:

x = [ log(2) + 5 log(3) ] / [ 2 log(3) - 5 log(2) ].

We finally realize after staring at this expression for awhile that we can just evaluate it using our calculator. So we are essentially done.

You should evaluate this expression using your calculator. Watch your signs of grouping and be careful. You should get -4.876747.

9.  Which of the following are valid and which are not?

log(x^y) = x log(y)

log(x/y) = log(x) - log(y)

log (x * y) = log(x) * log(y)

2 log(x) = log(2x)

log(x + y) = log(x) + log(y)

log(x) + log(y) = log(xy)

log(x^y) = (log(x)) ^ y

log(x - y) = log(x) - log(y)

3 log(x) = log(x^3)

log(x^y) = y + log(x)

log(x/y) = log(x) / log(y)

log(x^y) = y log(x)

10.  Simplify the following, and evaluate where possible:

log {base 8} (1024)

log {base 2} (4 * 32) (note that 4 and 32 are both powers of the base 2)

log (1000) ( recall that log without a base means base-10 logarithm)

ln(3xy)

log(3) + log(7) + log(41)

11.  Given that log(2) = .301, log(3) = .477, log(5) = .699, log(7) = .845, find log(4), log(6), log(8), and log(9). Use the given values to find the logarithm of every possible integer between 11 and 20.

12.  Solve the following equations, where possible, and state why it seems impossible to solve the rest:

2 ^ (5x) = 18

3 ^ (2x) = 7 ^ (x-4)

2^(3x) + 2^(4x) = 9

3^(2x-1) * 3^(3x+2) = 12

5x * 2^x = 9.