# 5.4: Logarithmic Functions - Mathematics

Learning Objectives

• Convert from logarithmic to exponential form.
• Convert from exponential to logarithmic form.
• Evaluate logarithms.
• Use common logarithms.
• Use natural logarithms.

In 2010, a major earthquake struck Haiti, destroying or damaging over 285,000 homes. One year later, another, stronger earthquake devastated Honshu, Japan, destroying or damaging over 332,000 buildings,like those shown in Figure (PageIndex{1}). Even though both caused substantial damage, the earthquake in 2011 was 100 times stronger than the earthquake in Haiti. How do we know? The magnitudes of earthquakes are measured on a scale known as the Richter Scale. The Haitian earthquake registered a 7.0 on the Richter Scalewhereas the Japanese earthquake registered a 9.0. The Richter Scale is a base-ten logarithmic scale. In other words, an earthquake of magnitude (8) is not twice as great as an earthquake of magnitude (4). It is

[10^{8−4}=10^4=10,000 onumber]

times as great! In this lesson, we will investigate the nature of the Richter Scale and the base-ten function upon which it depends.

## Converting from Logarithmic to Exponential Form

In order to analyze the magnitude of earthquakes or compare the magnitudes of two different earthquakes, we need to be able to convert between logarithmic and exponential form. For example, suppose the amount of energy released from one earthquake were 500 times greater than the amount of energy released from another. We want to calculate the difference in magnitude. The equation that represents this problem is (10^x=500), where (x) represents the difference in magnitudes on the Richter Scale. How would we solve for (x)?

We have not yet learned a method for solving exponential equations. None of the algebraic tools discussed so far is sufficient to solve (10^x=500). We know that ({10}^2=100) and ({10}^3=1000), so it is clear that (x) must be some value between 2 and 3, since (y={10}^x) is increasing. We can examine a graph, as in Figure (PageIndex{1}), to better estimate the solution. Estimating from a graph, however, is imprecise. To find an algebraic solution, we must introduce a new function. Observe that the graph in Figure (PageIndex{2}) passes the horizontal line test. The exponential function (y=b^x) is one-to-one, so its inverse, (x=b^y) is also a function. As is the case with all inverse functions, we simply interchange (x) and (y) and solve for (y) to find the inverse function. To represent (y) as a function of (x), we use a logarithmic function of the form (y={log}_b(x)). The base (b) logarithm of a number is the exponent by which we must raise (b) to get that number.

We read a logarithmic expression as, “The logarithm with base (b) of (x) is equal to (y),” or, simplified, “log base (b) of (x) is (y).” We can also say, “(b) raised to the power of (y) is (x),” because logs are exponents. For example, the base (2) logarithm of (32) is (5), because (5) is the exponent we must apply to (2) to get (32). Since (2^5=32), we can write ({log}_232=5). We read this as “log base (2) of (32) is (5).”

We can express the relationship between logarithmic form and its corresponding exponential form as follows:

[egin{align} log_b(x)=yLeftrightarrow b^y=x, b> 0, b eq 1 end{align}]

Note that the base (b) is always positive. Because logarithm is a function, it is most correctly written as (log_b(x)), using parentheses to denote function evaluation, just as we would with (f(x)). However, when the input is a single variable or number, it is common to see the parentheses dropped and the expression written without parentheses, as (log_bx). Note that many calculators require parentheses around the (x).

We can illustrate the notation of logarithms as follows: Notice that, comparing the logarithm function and the exponential function, the input and the output are switched. This means (y=log_b(x)) and (y=b^x) are inverse functions.

DEFINITION OF THE LOGARITHMIC FUNCTION

A logarithm base (b) of a positive number (x) satisfies the following definition.

For (x>0), (b>0), (b≠1),

[egin{align} y={log}_b(x) ext{ is equivalent to } b^y=x end{align}]

where,

• we read ({log}_b(x)) as, “the logarithm with base (b) of (x)” or the “log base (b) of (x)."
• the logarithm (y) is the exponent to which (b) must be raised to get (x).

Also, since the logarithmic and exponential functions switch the (x) and (y) values, the domain and range of the exponential function are interchanged for the logarithmic function. Therefore,

• the domain of the logarithm function with base (b) is ((0,infty)).
• the range of the logarithm function with base (b) is ((−infty,infty)).

Q&A: Can we take the logarithm of a negative number?

No. Because the base of an exponential function is always positive, no power of that base can ever be negative. We can never take the logarithm of a negative number. Also, we cannot take the logarithm of zero. Calculators may output a log of a negative number when in complex mode, but the log of a negative number is not a real number.

How to: Given an equation in logarithmic form ({log}_b(x)=y), convert it to exponential form

1. Examine the equation (y={log}_bx) and identify (b), (y),and (x).
2. Rewrite ({log}_bx=y) as (b^y=x).

Example (PageIndex{1}): Converting from Logarithmic Form to Exponential Form​​​​​​

Write the following logarithmic equations in exponential form.

1. ({log}_6(sqrt{6})=dfrac{1}{2})
2. ({log}_3(9)=2)

Solution

First, identify the values of (b), (y),and (x). Then, write the equation in the form (b^y=x).

1. ({log}_6(sqrt{6})=dfrac{1}{2})

Here, (b=6), (y=dfrac{1}{2}),and (x=sqrt{6}). Therefore, the equation ({log}_6(sqrt{6})=dfrac{1}{2}) is equivalent to

(6^{ frac{1}{2}}=sqrt{6})

2. ({log}_3(9)=2)

Here, (b=3), (y=2),and (x=9). Therefore, the equation ({log}_3(9)=2) is equivalent to

(3^2=9)

Exercise (PageIndex{1})

Write the following logarithmic equations in exponential form.

1. ({log}_{10}(1,000,000)=6)
2. ({log}_5(25)=2)

({log}_{10}(1,000,000)=6) is equivalent to ({10}^6=1,000,000)

({log}_5(25)=2) is equivalent to (5^2=25)

## Converting from Exponential to Logarithmic Form

To convert from exponents to logarithms, we follow the same steps in reverse. We identify the base (b),exponent (x),and output (y). Then we write (x={log}_b(y)).

Example (PageIndex{2}): Converting from Exponential Form to Logarithmic Form

Write the following exponential equations in logarithmic form.

1. (2^3=8)
2. (5^2=25)
3. ({10}^{−4}=dfrac{1}{10,000})

Solution

First, identify the values of (b), (y),and (x). Then, write the equation in the form (x={log}_b(y)).

1. (2^3=8)

Here, (b=2), (x=3),and (y=8). Therefore, the equation (2^3=8) is equivalent to ({log}_2(8)=3).

2. (5^2=25)

Here, (b=5), (x=2),and (y=25). Therefore, the equation (5^2=25) is equivalent to ({log}_5(25)=2).

3. ({10}^{−4}=dfrac{1}{10,000})

Here, (b=10), (x=−4),and (y=dfrac{1}{10,000}). Therefore, the equation ({10}^{−4}=dfrac{1}{10,000}) is equivalent to ({log}_{10} left (dfrac{1}{10,000} ight )=−4).

Exercise (PageIndex{2})

Write the following exponential equations in logarithmic form.

1. (3^2=9)
2. (5^3=125)
3. (2^{−1}=dfrac{1}{2})

(3^2=9) is equivalent to ({log}_3(9)=2)

(5^3=125) is equivalent to ({log}_5(125)=3)

(2^{−1}=dfrac{1}{2}) is equivalent to ({log}_2 left (dfrac{1}{2} ight )=−1)

## Evaluating Logarithms

Knowing the squares, cubes, and roots of numbers allows us to evaluate many logarithms mentally. For example, consider ({log}_28). We ask, “To what exponent must (2) be raised in order to get 8?” Because we already know (2^3=8), it follows that ({log}_28=3).

Now consider solving ({log}_749) and ({log}_327) mentally.

• We ask, “To what exponent must (7) be raised in order to get (49)?” We know (7^2=49). Therefore, ({log}_749=2)
• We ask, “To what exponent must (3) be raised in order to get (27)?” We know (3^3=27). Therefore, (log_{3}27=3)

Even some seemingly more complicated logarithms can be evaluated without a calculator. For example, let’s evaluate (log_{ce{2/3}} frac{4}{9}) mentally.

• We ask, “To what exponent must (ce{2/3}) be raised in order to get (ce{4/9})? ” We know (2^2=4) and (3^2=9), so [{left(dfrac{2}{3} ight )}^2=dfrac{4}{9}. onumber] Therefore, [{log}_{ce{2/3}} left (dfrac{4}{9} ight )=2. onumber]

How to: Given a logarithm of the form (y={log}_b(x)),evaluate it mentally

1. Rewrite the argument (x) as a power of (b): (b^y=x).
2. Use previous knowledge of powers of (b) identify (y) by asking, “To what exponent should (b) be raised in order to get (x)?”

Example (PageIndex{3}): Solving Logarithms Mentally

Solve (y={log}_4(64)) without using a calculator.

Solution

First we rewrite the logarithm in exponential form: (4^y=64). Next, we ask, “To what exponent must (4) be raised in order to get (64)?”

We know

(4^3=64)

Therefore,

({log}_4(64)=3)

Exercise (PageIndex{3})

Solve (y={log}_{121}(11)) without using a calculator.

({log}_{121}(11)=dfrac{1}{2}) (recalling that (sqrt{121}={(121)}^{ frac{1}{2}}=11))

Example (PageIndex{4}): Evaluating the Logarithm of a Reciprocal

Evaluate (y={log}_3 left (dfrac{1}{27} ight )) without using a calculator.

Solution

First we rewrite the logarithm in exponential form: (3^y=dfrac{1}{27}). Next, we ask, “To what exponent must (3) be raised in order to get (dfrac{1}{27})?”

We know (3^3=27),but what must we do to get the reciprocal, (dfrac{1}{27})? Recall from working with exponents that (b^{−a}=dfrac{1}{b^a}). We use this information to write

[egin{align*} 3^{-3}&= dfrac{1}{3^3} &= dfrac{1}{27} end{align*}]

Therefore, ({log}_3 left (dfrac{1}{27} ight )=−3).

Exercise (PageIndex{4})

Evaluate (y={log}_2 left (dfrac{1}{32} ight )) without using a calculator.

({log}_2 left (dfrac{1}{32} ight )=−5)

## Using Common Logarithms

Sometimes we may see a logarithm written without a base. In this case, we assume that the base is (10). In other words, the expression (log(x)) means ({log}_{10}(x)). We call a base (-10) logarithm a common logarithm. Common logarithms are used to measure the Richter Scale mentioned at the beginning of the section. Scales for measuring the brightness of stars and the pH of acids and bases also use common logarithms.

DEFINITION OF THE COMMON LOGARITHM

A common logarithm is a logarithm with base (10). We write ({log}_{10}(x)) simply as (log(x)). The common logarithm of a positive number (x) satisfies the following definition.

For (x>0),

[egin{align} y={log}(x) ext{ is equivalent to } {10}^y=x end{align}]

We read (log(x)) as, “the logarithm with base (10) of (x) ” or “log base (10) of (x).”

The logarithm (y) is the exponent to which (10) must be raised to get (x).

How to: Given a common logarithm of the form (y=log(x)), evaluate it mentally

1. Rewrite the argument (x) as a power of (10): ({10}^y=x).
2. Use previous knowledge of powers of (10) to identify (y) by asking, “To what exponent must (10) be raised in order to get (x)?”

Example (PageIndex{5}): Finding the Value of a Common Logarithm Mentally

Evaluate (y=log(1000)) without using a calculator.

Solution

First we rewrite the logarithm in exponential form: ({10}^y=1000). Next, we ask, “To what exponent must (10) be raised in order to get (1000)?” We know

({10}^3=1000)

Therefore, (log(1000)=3).

Exercise (PageIndex{5})

Evaluate (y=log(1,000,000)).

(log(1,000,000)=6)

How to: Given a common logarithm with the form (y=log(x)),evaluate it using a calculator

1. Press [LOG].
2. Enter the value given for (x),followed by [ ) ].
3. Press [ENTER].

Example (PageIndex{6}): ​​​​​​Finding the Value of a Common Logarithm Using a Calculator

Evaluate (y=log(321)) to four decimal places using a calculator.

Solution

• Press [LOG].
• Enter 321, followed by [ ) ].
• Press [ENTER].

Rounding to four decimal places, (log(321)≈2.5065).

Analysis

Note that ({10}^2=100) and that ({10}^3=1000). Since (321) is between (100) and (1000), we know that (log(321)) must be between (log(100)) and (log(1000)). This gives us the following:

(100<321<1000)

(2<2.5065<3)

Exercise (PageIndex{6})

Evaluate (y=log(123)) to four decimal places using a calculator.

(log(123)≈2.0899)

Example (PageIndex{7}): Rewriting and Solving a Real-World Exponential Model

The amount of energy released from one earthquake was (500) times greater than the amount of energy released from another. The equation ({10}^x=500) represents this situation, where (x) is the difference in magnitudes on the Richter Scale. To the nearest thousandth, what was the difference in magnitudes?

Solution

We begin by rewriting the exponential equation in logarithmic form.

({10}^x=500)

(log(500)=x) Use the definition of the common log.

Next we evaluate the logarithm using a calculator:

• Press [LOG].
• Enter (500),followed by [ ) ].
• Press [ENTER].
• To the nearest thousandth, (log(500)≈2.699).

The difference in magnitudes was about (2.699).

Exercise (PageIndex{7})

​​​​The amount of energy released from one earthquake was (8,500) times greater than the amount of energy released from another. The equation ({10}^x=8500) represents this situation, where (x) is the difference in magnitudes on the Richter Scale. To the nearest thousandth, what was the difference in magnitudes?

The difference in magnitudes was about (3.929).

## Using Natural Logarithms

The most frequently used base for logarithms is (e). Base (e) logarithms are important in calculus and some scientific applications; they are called natural logarithms. The base (e) logarithm, ({log}_e(x)), has its own notation,(ln(x)). Most values of (ln(x)) can be found only using a calculator. The major exception is that, because the logarithm of (1) is always (0) in any base, (ln1=0). For other natural logarithms, we can use the (ln) key that can be found on most scientific calculators. We can also find the natural logarithm of any power of (e) using the inverse property of logarithms.

DEFINITION OF THE NATURAL LOGARITHM

A natural logarithm is a logarithm with base (e). We write ({log}_e(x)) simply as (ln(x)). The natural logarithm of a positive number (x) satisfies the following definition.

For (x>0),

(y=ln(x)) is equivalent to (e^y=x)

We read (ln(x)) as, “the logarithm with base (e) of (x)” or “the natural logarithm of (x).”

The logarithm (y) is the exponent to which (e) must be raised to get (x).

Since the functions (y=e^x) and (y=ln(x)) are inverse functions, (ln(e^x)=x) for all (x) and (e^{ln (x)}=x) for (x>0).

How to: Given a natural logarithm with the form (y=ln(x)), evaluate it using a calculator

1. Press [LN].
2. Enter the value given for (x), followed by [ ) ].
3. Press [ENTER].

Example (PageIndex{8}): Evaluating a Natural Logarithm Using a Calculator

Evaluate (y=ln(500)) to four decimal places using a calculator.

Solution

• Press [LN].
• Enter (500),followed by [ ) ].
• Press [ENTER].

Rounding to four decimal places, (ln(500)≈6.2146)

Exercise (PageIndex{8})

Evaluate (ln(−500)).

It is not possible to take the logarithm of a negative number in the set of real numbers.

Media

Access this online resource for additional instruction and practice with logarithms.

• Introduction to Logarithms

## Key Equations

 Definition of the logarithmic function For (x>0), (b>0), (b≠1), (y={log}_b(x)) if and only if (b^y=x). Definition of the common logarithm For (x>0), (y=log(x)) if and only if ({10}^y=x). Definition of the natural logarithm For (x>0), (y=ln(x)) if and only if (e^y=x).

## Key Concepts

• The inverse of an exponential function is a logarithmic function, and the inverse of a logarithmic function is an exponential function.
• Logarithmic equations can be written in an equivalent exponential form, using the definition of a logarithm. See Example (PageIndex{1}).
• Exponential equations can be written in their equivalent logarithmic form using the definition of a logarithm See Example (PageIndex{2}).
• Logarithmic functions with base (b) can be evaluated mentally using previous knowledge of powers of (b). See Example (PageIndex{3}) and Example (PageIndex{4}).
• Common logarithms can be evaluated mentally using previous knowledge of powers of (10). See Example (PageIndex{5}).
• When common logarithms cannot be evaluated mentally, a calculator can be used. See Example (PageIndex{6}).
• Real-world exponential problems with base (10) can be rewritten as a common logarithm and then evaluated using a calculator. See Example (PageIndex{7}).
• Natural logarithms can be evaluated using a calculator Example (PageIndex{8}).

## 5.4: Logarithmic Functions - Mathematics

For problems 1 – 3 write the expression in logarithmic form.

For problems 4 – 6 write the expression in exponential form.

For problems 7 - 12 determine the exact value of each of the following without using a calculator.

1. (81) Solution
2. (125) Solution
3. (displaystyle frac<1><8>) Solution
4. (<4>>>16) Solution
5. (ln <<f>^4>) Solution
6. (log displaystyle frac<1><<100>>) Solution

For problems 13 – 15 write each of the following in terms of simpler logarithms

For problems 16 – 18 combine each of the following into a single logarithm with a coefficient of one.

1. (2x + 5y - frac<1><2>z) Solution
2. (3ln left( ight) - 4ln t - 2ln left( ight)) Solution
3. (displaystyle frac<1><3>log a - 6log b + 2) Solution

For problems 19 & 20 use the change of base formula and a calculator to find the value of each of the following.

## 5.4: Logarithmic Functions - Mathematics

In this section we’ll take a look at a function that is related to the exponential functions we looked at in the last section. We will look at logarithms in this section. Logarithms are one of the functions that students fear the most. The main reason for this seems to be that they simply have never really had to work with them. Once they start working with them, students come to realize that they aren’t as bad as they first thought.

We’ll start with (b > 0), (b e 1) just as we did in the last section. Then we have

The first is called logarithmic form and the second is called the exponential form. Remembering this equivalence is the key to evaluating logarithms. The number, (b), is called the base.

Let's do some quick evaluations.

To quickly evaluate logarithms the easiest thing to do is to convert the logarithm to exponential form. So, let’s take a look at the first one.

First, let’s convert to exponential form.

So, we’re really asking 2 raised to what gives 16. Since 2 raised to 4 is 16 we get,

We’ll not do the remaining parts in quite this detail, but they were all work in this way.

Note the difference between the first and second logarithm! The base is important! It can completely change the answer.

There are a couple of special logarithms that arise in many places. These are,

In the natural logarithm the base e is the same number as in the natural exponential logarithm that we saw in the last section. Here is a sketch of both of these logarithms. From this graph we can get a couple of very nice properties about the natural logarithm that we will use many times in this and later Calculus courses.

[eginln x o infty >x o infty ln x o - infty >x o 0,,,x > 0end]

Let’s take a look at a couple of more logarithm evaluations. Some of which deal with the natural or common logarithm and some of which don’t.

These work exactly the same as previous example so we won’t put in too many details.

This last set of examples leads us to some of the basic properties of logarithms.

#### Properties

1. The domain of the logarithm function is (left( <0,infty > ight)). In other words, we can only plug positive numbers into a logarithm! We can’t plug in zero or a negative number.
2. The range of the logarithm function is (left( < - infty ,infty > ight)).
3. (b = 1)
4. (1 = 0)
5. (= x)
6. (_b>x>> = x)

The last two properties will be especially useful in the next section. Notice as well that these last two properties tell us that,

Here are some more properties that are useful in the manipulation of logarithms.

#### More Properties

Note that there is no equivalent property to the first two for sums and differences. In other words,

What the instructions really mean here is to use as many of the properties of logarithms as we can to simplify things down as much as we can.

Property 7 above can be extended to products of more than two functions. Once we’ve used Property 7 we can then use Property 9.

When using property 8 above make sure that the logarithm that you subtract is the one that contains the denominator as its argument. Also, note that that we’ll be converting the root to fractional exponents in the first step.

The point to this problem is mostly the correct use of property 9 above.

You can use Property 9 on the second term because the WHOLE term was raised to the 3, but in the first logarithm, only the individual terms were squared and not the term as a whole so the 2’s must stay where they are!

The last topic that we need to look at in this section is the change of base formula for logarithms. The change of base formula is,

This is the most general change of base formula and will convert from base (b) to base (a). However, the usual reason for using the change of base formula is to compute the value of a logarithm that is in a base that you can’t easily deal with. Using the change of base formula means that you can write the logarithm in terms of a logarithm that you can deal with. The two most common change of base formulas are

In fact, often you will see one or the other listed as THE change of base formula!

In the first part of this section we computed the value of a few logarithms, but we could do these without the change of base formula because all the arguments could be written in terms of the base to a power. For instance,

However, this only works because 49 can be written as a power of 7! We would need the change of base formula to compute (50).

So, it doesn’t matter which we use, we will get the same answer regardless of the logarithm that we use in the change of base formula.

Note as well that we could use the change of base formula on (49) if we wanted to as well.

This is a lot of work however, and is probably not the best way to deal with this.

So, in this section we saw how logarithms work and took a look at some of the properties of logarithms. We will run into logarithms on occasion so make sure that you can deal with them when we do run into them.

## 5.4: Logarithmic Functions - Mathematics

In this section we now need to move into logarithm functions. This can be a tricky function to graph right away. There is going to be some different notation that you aren’t used to and some of the properties may not be all that intuitive. Do not get discouraged however. Once you figure these out you will find that they really aren’t that bad and it usually just takes a little working with them to get them figured out.

Here is the definition of the logarithm function.

If (b) is any number such that (b > 0) and (b e 1) and (x > 0) then,

We usually read this as “log base (b) of (x)”.

In this definition (y = x) is called the logarithm form and ( = x) is called the exponential form.

Note that the requirement that (x > 0) is really a result of the fact that we are also requiring (b > 0). If you think about it, it will make sense. We are raising a positive number to an exponent and so there is no way that the result can possibly be anything other than another positive number. It is very important to remember that we can’t take the logarithm of zero or a negative number.

Now, let’s address the notation used here as that is usually the biggest hurdle that students need to overcome before starting to understand logarithms. First, the “log” part of the function is simply three letters that are used to denote the fact that we are dealing with a logarithm. They are not variables and they aren’t signifying multiplication. They are just there to tell us we are dealing with a logarithm.

Next, the (b) that is subscripted on the “log” part is there to tell us what the base is as this is an important piece of information. Also, despite what it might look like there is no exponentiation in the logarithm form above. It might look like we’ve got () in that form, but it isn’t. It just looks like that might be what’s happening.

It is important to keep the notation with logarithms straight, if you don’t you will find it very difficult to understand them and to work with them.

Now, let’s take a quick look at how we evaluate logarithms.

Now, the reality is that evaluating logarithms directly can be a very difficult process, even for those who really understand them. It is usually much easier to first convert the logarithm form into exponential form. In that form we can usually get the answer pretty quickly.

Okay what we are really asking here is the following.

As suggested above, let’s convert this to exponential form.

Most people cannot evaluate the logarithm (16) right off the top of their head. However, most people can determine the exponent that we need on 4 to get 16 once we do the exponentiation. So, since,

we must have the following value of the logarithm.

This one is similar to the previous part. Let’s first convert to exponential form.

If you don’t know this answer right off the top of your head, start trying numbers. In other words, compute (<2^2>), (<2^3>), (<2^4>), etc until you get 16. In this case we need an exponent of 4. Therefore, the value of this logarithm is,

Before moving on to the next part notice that the base on these is a very important piece of notation. Changing the base will change the answer and so we always need to keep track of the base.

We’ll do this one without any real explanation to see how well you’ve got the evaluation of logarithms down.

Now, this one looks different from the previous parts, but it really isn’t any different. As always let’s first convert to exponential form.

First, notice that the only way that we can raise an integer to an integer power and get a fraction as an answer is for the exponent to be negative. So, we know that the exponent has to be negative.

Now, let’s ignore the fraction for a second and ask ( <5^?>= 125). In this case if we cube 5 we will get 125.

So, it looks like we have the following,

Converting this logarithm to exponential form gives,

Now, just like the previous part, the only way that this is going to work out is if the exponent is negative. Then all we need to do is recognize that ( <3^4>= 81) and we can see that,

Here is the answer to this one.

Hopefully, you now have an idea on how to evaluate logarithms and are starting to get a grasp on the notation. There are a few more evaluations that we want to do however, we need to introduce some special logarithms that occur on a very regular basis. They are the common logarithm and the natural logarithm. Here are the definitions and notations that we will be using for these two logarithms.

So, the common logarithm is simply the log base 10, except we drop the “base 10” part of the notation. Similarly, the natural logarithm is simply the log base (f) with a different notation and where (f) is the same number that we saw in the previous section and is defined to be (<f> = 2.718281828 ldots ).

Let’s take a look at a couple more evaluations.

1. (log 1000)
2. (log displaystyle frac<1><<100>>)
3. (ln displaystyle frac<1><<f>>)
4. (ln sqrt <f> )
5. (>34)
6. (1)

To do the first four evaluations we just need to remember what the notation for these are and what base is implied by the notation. The final two evaluations are to illustrate some of the properties of all logarithms that we’ll be looking at eventually.

(log 1000 = 3) because ( <10^3>= 1000).

(ln sqrt <f> = frac<1><2>) because (<<f>^<2>>> = sqrt <f> ). Notice that with this one we are really just acknowledging a change of notation from fractional exponent into radical form.

(>34 = 1) because ( <34^1>= 34). Notice that this one will work regardless of the base that we’re using.

(1 = 0) because ( <8^0>= 1). Again, note that the base that we’re using here won’t change the answer.

So, when evaluating logarithms all that we’re really asking is what exponent did we put onto the base to get the number in the logarithm.

Now, before we get into some of the properties of logarithms let’s first do a couple of quick graphs.

This example has two points. First, it will familiarize us with the graphs of the two logarithms that we are most likely to see in other classes. Also, it will give us some practice using our calculator to evaluate these logarithms because the reality is that is how we will need to do most of these evaluations.

Here is a table of values for the two logarithms.

(x) (log x) (ln x)
(frac<1><2>) -0.3010 -0.6931
1 0 0
2 0.3010 0.6931
3 0.4771 1.0986
4 0.6021 1.3863

Here is a sketch of the graphs of these two functions. Now let’s start looking at some properties of logarithms. We’ll start off with some basic evaluation properties.

#### Properties of Logarithms

1. (1 = 0). This follows from the fact that ( = 1).

Properties 3 and 4 leads to a nice relationship between the logarithm and exponential function. Let’s first compute the following function compositions for (fleft( x ight) = ) and (gleft( x ight) = x).

Recall from the section on inverse functions that this means that the exponential and logarithm functions are inverses of each other. This is a nice fact to remember on occasion.

We should also give the generalized version of Properties 3 and 4 in terms of both the natural and common logarithm as we’ll be seeing those in the next couple of sections on occasion.

Now, let’s take a look at some manipulation properties of the logarithm.

#### More Properties of Logarithms

We won’t be doing anything with the final property in this section it is here only for the sake of completeness. We will be looking at this property in detail in a couple of sections.

The first two properties listed here can be a little confusing at first since on one side we’ve got a product or a quotient inside the logarithm and on the other side we’ve got a sum or difference of two logarithms. We will just need to be careful with these properties and make sure to use them correctly.

Also, note that there are no rules on how to break up the logarithm of the sum or difference of two terms. To be clear about this let’s note the following,

Be careful with these and do not try to use these as they simply aren’t true.

Note that all of the properties given to this point are valid for both the common and natural logarithms. We just didn’t write them out explicitly using the notation for these two logarithms, the properties do hold for them nonetheless

Now, let’s see some examples of how to use these properties.

The instructions here may be a little misleading. When we say simplify we really mean to say that we want to use as many of the logarithm properties as we can.

Note that we can’t use Property 7 to bring the 3 and the 5 down into the front of the logarithm at this point. In order to use Property 7 the whole term in the logarithm needs to be raised to the power. In this case the two exponents are only on individual terms in the logarithm and so Property 7 can’t be used here.

We do, however, have a product inside the logarithm so we can use Property 5 on this logarithm.

Now that we’ve done this we can use Property 7 on each of these individual logarithms to get the final simplified answer.

In this case we’ve got a product and a quotient in the logarithm. In these cases it is almost always best to deal with the quotient before dealing with the product. Here is the first step in this part.

Now, we’ll break up the product in the first term and once we’ve done that we’ll take care of the exponents on the terms.

For this part let’s first rewrite the logarithm a little so that we can see the first step.

Written in this form we can see that there is a single exponent on the whole term and so we’ll take care of that first.

Now, we will take care of the product.

Notice the parenthesis in this the answer. The (frac<1><2>) multiplies the original logarithm and so it will also need to multiply the whole “simplified” logarithm. Therefore, we need to have a set of parenthesis there to make sure that this is taken care of correctly.

We’ll first take care of the quotient in this logarithm.

We now reach the real point to this problem. The second logarithm is as simplified as we can make it. Remember that we can’t break up a log of a sum or difference and so this can’t be broken up any farther. Also, we can only deal with exponents if the term as a whole is raised to the exponent. The fact that both pieces of this term are squared doesn’t matter. It needs to be the whole term squared, as in the first logarithm.

So, we can further simplify the first logarithm, but the second logarithm can’t be simplified any more. Here is the final answer for this problem.

Now, we need to work some examples that go the other way. This next set of examples is probably more important than the previous set. We will be doing this kind of logarithm work in a couple of sections.

The instruction requiring a coefficient of 1 means that the when we get down to a final logarithm there shouldn’t be any number in front of the logarithm.

Note as well that these examples are going to be using Properties 5 – 7 only we’ll be using them in reverse. We will have expressions that look like the right side of the property and use the property to write it so it looks like the left side of the property.

The first step here is to get rid of the coefficients on the logarithms. This will use Property 7 in reverse. In this direction, Property 7 says that we can move the coefficient of a logarithm up to become a power on the term inside the logarithm.

Here is that step for this part.

We’ve now got a sum of two logarithms both with coefficients of 1 and both with the same base. This means that we can use Property 5 in reverse. Here is the answer for this part.

Again, we will first take care of the coefficients on the logarithms.

We now have a difference of two logarithms and so we can use Property 6 in reverse. When using Property 6 in reverse remember that the term from the logarithm that is subtracted off goes in the denominator of the quotient. Here is the answer to this part.

In this case we’ve got three terms to deal with and none of the properties have three terms in them. That isn’t a problem. Let’s first take care of the coefficients and at the same time we’ll factor a minus sign out of the last two terms. The reason for this will be apparent in the next step.

[5ln left( ight) - 2ln y - 8ln x = ln ight)^5> - left( + ln > ight)]

Now, notice that the quantity in the parenthesis is a sum of two logarithms and so can be combined into a single logarithm with a product as follows,

[5ln left( ight) - 2ln y - 8ln x = ln ight)^5> - ln left( <> ight)]

Now we are down to two logarithms and they are a difference of logarithms and so we can write it as a single logarithm with a quotient.

The final topic that we need to discuss in this section is the change of base formula.

Most calculators these days are capable of evaluating common logarithms and natural logarithms. However, that is about it, so what do we do if we need to evaluate another logarithm that can’t be done easily as we did in the first set of examples that we looked at?

To do this we have the change of base formula. Here is the change of base formula.

where we can choose (b) to be anything we want it to be. In order to use this to help us evaluate logarithms this is usually the common or natural logarithm. Here is the change of base formula using both the common logarithm and the natural logarithm.

Let’s see how this works with an example.

First, notice that we can’t use the same method to do this evaluation that we did in the first set of examples. This would require us to look at the following exponential form,

and that’s just not something that anyone can answer off the top of their head. If the 7 had been a 5, or a 25, or a 125, etc. we could do this, but it’s not. Therefore, we have to use the change of base formula.

Now, we can use either one and we’ll get the same answer. So, let’s use both and verify that. We’ll start with the common logarithm form of the change of base.

## 6.5 Logarithmic Properties

In chemistry, pH is used as a measure of the acidity or alkalinity of a substance. The pH scale runs from 0 to 14. Substances with a pH less than 7 are considered acidic, and substances with a pH greater than 7 are said to be alkaline. Our bodies, for instance, must maintain a pH close to 7.35 in order for enzymes to work properly. To get a feel for what is acidic and what is alkaline, consider the following pH levels of some common substances:

• Battery acid: 0.8
• Stomach acid: 2.7
• Orange juice: 3.3
• Pure water: 7 (at 25° C)
• Human blood: 7.35
• Fresh coconut: 7.8
• Sodium hydroxide (lye): 14

To determine whether a solution is acidic or alkaline, we find its pH, which is a measure of the number of active positive hydrogen ions in the solution. The pH is defined by the following formula, where H + H + is the concentration of hydrogen ion in the solution

### Using the Product Rule for Logarithms

Recall that the logarithmic and exponential functions “undo” each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here. First, the following properties are easy to prove.

## Properties of Logarithms

One of the powerful things about Logarithms is that they can turn multiply into add.

"the log of multiplication is the sum of the logs"

### Why is that true? See Footnote.

Using that property and the Laws of Exponents we get these useful properties:

 loga(m × n) = logam + logan the log of multiplication is the sum of the logs loga(m/n) = logam &minus logan the log of division is the difference of the logs loga(1/n) = &minuslogan this just follows on from the previous "division" rule, because loga(1) = 0 loga(m r ) = r ( logam ) the log of m with an exponent r is r times the log of m

Remember: the base "a" is always the same! History: Logarithms were very useful before calculators were invented . for example, instead of multiplying two large numbers, by using logarithms you could turn it into addition (much easier!)

## 5.4: Logarithmic Functions - Mathematics

Introduction to Logarithmic Functions

· Convert logarithmic equations to exponential equations.

· Convert exponential equations to logarithmic equations.

· Graph logarithmic functions.

· Solve logarithmic equations.

A useful family of functions that is related to exponential functions is the logarithmic functions. You have been calculating the result of b x , and this gave us the exponential functions. A logarithm is a calculation of the exponent in the equation y = b x . Put another way, finding a logarithm is the same as finding the exponent to which the given base must be raised to get the desired value. The exponent becomes the output rather than the input.

Calculating Exponents

Consider these tables of values using a base of 2.

Input x, an exponent

Input x, a number that is a power of 2.

Output y, the exponent of 2.      Note the two tables are the same except the columns are reversed—the point (1, 2) taken from the first table will be the point (2, 1) in the second table.

The graphs of these two relationships should have the same general shape. As shown in the graph, the two curves are symmetrical about the line y = x. Another way to put it, if you rotate the red curve about the line y = x, it will coincide with the blue curve. (This makes sense, because y in the first table becomes x in the second table, and vice versa.) The equation x = 2 y is often written as a logarithmic function (called log function for short). The logarithmic function for x = 2 y is written as y = log2 x or f(x) = log2 x. The number 2 is still called the base. In general, y = logb x is read, “y equals log to the base b of x,” or more simply, “y equals log base b of x.” As with exponential functions, b > 0 and b ≠ 1.

You can see from the graph that the range (y values) of the exponential function (in red) is positive real numbers. Since the input and output have been switched, the domain (x values) of the logarithmic function (in blue) is positive real numbers.

Similarly, the domain of the exponential function (in red) is all real numbers. The range of the logarithmic function (in blue) is all real numbers.

Definition of Logarithm

The logarithm of x in base b is written logb x and is defined as:

logb x = y if and only if b y = x, where x > 0 and b > 0, b ≠ 1.

Connecting Exponential and Logarithmic Equations

It’s important to remember that the result of a logarithm is the exponent! That is, logb x asks, “What exponent on the base b will give the result x?” Sometimes, you will need to convert logb x = y to b y = x. Other times, you will convert b y = x to logb x = y. The examples in the table show the logarithmic form and the corresponding exponential form of several equations.

## 6.3 Logarithmic Functions

In 2010, a major earthquake struck Haiti, destroying or damaging over 285,000 homes 19 . One year later, another, stronger earthquake devastated Honshu, Japan, destroying or damaging over 332,000 buildings, 20 like those shown in Figure 1. Even though both caused substantial damage, the earthquake in 2011 was 100 times stronger than the earthquake in Haiti. How do we know? The magnitudes of earthquakes are measured on a scale known as the Richter Scale. The Haitian earthquake registered a 7.0 on the Richter Scale 21 whereas the Japanese earthquake registered a 9.0. 22

The Richter Scale is a base-ten logarithmic scale. In other words, an earthquake of magnitude 8 is not twice as great as an earthquake of magnitude 4. It is 10 8 − 4 = 10 4 = 10,000 10 8 − 4 = 10 4 = 10,000 times as great! In this lesson, we will investigate the nature of the Richter Scale and the base-ten function upon which it depends.

### Converting from Logarithmic to Exponential Form

In order to analyze the magnitude of earthquakes or compare the magnitudes of two different earthquakes, we need to be able to convert between logarithmic and exponential form. For example, suppose the amount of energy released from one earthquake were 500 times greater than the amount of energy released from another. We want to calculate the difference in magnitude. The equation that represents this problem is 10 x = 500 , 10 x = 500 , where x x represents the difference in magnitudes on the Richter Scale . How would we solve for x ? x ?

We have not yet learned a method for solving exponential equations. None of the algebraic tools discussed so far is sufficient to solve 10 x = 500. 10 x = 500. We know that 10 2 = 100 10 2 = 100 and 10 3 = 1000 , 10 3 = 1000 , so it is clear that x x must be some value between 2 and 3, since y = 10 x y = 10 x is increasing. We can examine a graph, as in Figure 2, to better estimate the solution.

We can express the relationship between logarithmic form and its corresponding exponential form as follows:

We can illustrate the notation of logarithms as follows:

Notice that, comparing the logarithm function and the exponential function, the input and the output are switched. This means y = log b ( x ) y = log b ( x ) and y = b x y = b x are inverse functions.

## SOLVING LOGARITHMIC EQUATIONS

Solution: Step 1: As we know by now, we can only take the logarithm of a positive number. Therefore, we need to make a restriction on the domain (values of x) so that the problem will be valid (have an answer).
The term is valid when x + 5 >0 or x > -5 the term is valid when x + 2 > 0 or x > -2 and the term is valid when x + 6 > 0 or x > -6. If we require that the domain be restricted to the set of all real numbers such that x > -2, all the terms will be valid. Step 2: Simplify the left side of the original equation using Logarithmic Rule 1:

Step 3: We now have an equation of the form which implies that the expression a must equal the expression b or:

Step 4: You could also raise the base 2 to an exponent equal to the left side of the equation in Step 2, and you could raise the base 2 to an exponent equal to right side of the equation in Step 2:

When the base is the same as the base of the logarithm, the above equation can be simplified to

Step 5: Simplify the left side of the above equation:

Step 6: Subtract x and subtract 6 from both sides of the above equation:

Step 7: Use the quadratic formula to solve for x:

There are two exact answer: and and there are two approximate answers:

However only one of the answers is valid.
One of the answers ( ) is out of our specified domain of the set of all real numbers such that (x > -2).
Therefore, the exact and approximate answers are:

Check: Let's substitute the value in the original equation and determine whether the left side of the equation equals the right side of the equation after the substitution. In other words, does

The answer checks. We can also check our solution with the approximate answer. Does

Since the value of the left side of the original equation equals the right side of the original equation when we substitute the exact and the approximate value of x, We have proved the answer.
Let's illustrate why we had to throw out one of the answers. Let's check to see if the approximate solution (the one we discarded) works:

At this point we have to stop because we cannot take the log of a negative number. We simply cannot calculate the value if any of the terms are undefined.

## Conclusion

Hopefully, whatever logarithmic graph you are trying to find the equation for will be covered by one or more of the cases above.

### Related posts:

Most people have seen some basic graphs before. Graphs are pictorial representations of data and.A reader asked how to find the equation of a parabola from its graph.Here's a satisfactory attempt to model the Earth's increase in CO2 concentration, using Excel and.A quintic curve is a polynomial of degree 5. Given such a curve, how.The Equation of Time is an interesting application of conics and composite trigonometric curves.

Posted in Mathematics category - 29 May 2019 [Permalink]

### 3 Comments on “How to find the equation of a logarithm function from its graph?”

The problem with using a logarithm (of time) to approximate the forgetting data is that it eventually predicts a negative proportion of correct answers.

@Alan Yes, I mentioned we usually use an exponential curve for forgetting data. I added a bit more of a caution just now.