3.4: Properties of Logarithms
Learning objectives
 Use properties of logarithms to write logarithmic expressions in different forms.
 Evaluate common logarithms and natural logarithms.
 Use the change of base formula and a scientific calculator to find the values of logs with any bases.
Introduction
In the previous lesson we defined the logarithmic function as the inverse of an exponential function, and we evaluated log expressions in order to identify values of these functions. In this lesson we will work with more complicated log expressions. We will develop properties of logs that we can use to write a log expression as the sum or difference of several expressions, or to write several expressions as a single log expression. We will also work with logs with base 10 and base e, which are the bases most often used in applications of logarithmic functions.
Properties of Logarithms
Because a logarithm is an exponent, the properties of logs are the same as the properties of exponents. Here we will prove several important properties of logarithms.
Property 1: log_{b}(xy) = log_{b} x + log_{b} y
Proof: Let log_{b} x = n and log_{b} y = m.
 Rewrite both log expressions in exponential form:

 log_{b} x = n → b^{n} = x

 log_{b} y = m → b^{m} = y
 Now multiply x and y: xy = b^{n} × b^{m} = b^{n + m}
 Therefore we have an exponential statement: b^{n + m} = xy.
 The log form of the statement is: log_{b} xy = n + m.
 Now recall how we defined n and m:



 log_{b} xy = n + m = log_{b} x + log_{b} y.


Property 2: \begin{align*}log_b \left (\frac{x} {y}\right ) = log_b x  log_b y\end{align*}
We can prove property 2 analogously to the way we proved property 1.
Proof: Let log_{b} x = n and log_{b} y = m.
 Rewrite both log expressions in exponential form:



 log_{b} x = n → b^{n} = x





 log_{b} y = m → b^{m} = y


 Now divide x by y: \begin{align*}\frac{x} {y} = \frac{b^n} {b^m} = b^{n  m}\end{align*}
 Therefore we have an exponential statement: \begin{align*}b^{n  m} = \frac{x} {y}\end{align*}.
 The log form of the statement is: \begin{align*}log_b \left (\frac{x} {y}\right ) = n  m\end{align*}.
 Now recall how we defined n and m:

 \begin{align*}log_b \left (\frac{x} {y}\right ) = n  m = log_b x  log_b y\end{align*}.
Property 3: log_{b} x^{n} = n log_{b} x
The proof of the third property relies on another property of logs that we can derive by thinking about the definition of a log. Consider the expression log_{2} 2^{13}. What does this expression mean?
The exponential form of log_{2} 2^{13} = ? is 2^{?} = 2^{13}. Looking at the exponential form should convince you that the missing exponent is 13. That is, log_{2} 2^{13} = 13. In general, log_{b} b^{n} = n. This property will be used in the proof of property 3.
Proof (of Property 3):
 Let log_{b} x = w.
 The exponential form of this log statement is b^{w} = x.
 If we raise both sides of this equation to the power of n, we have (b^{w})^{n} = x^{n}.
 Using the power property of exponents, this equation simplifies to b^{wn} = x^{n}
 If two expressions are equal, then the logs of both expressions are equal:



 log_{b} b^{wn} = log_{b} x^{n}


 Now consider the value of the left side of the equation: log_{b} b^{wn} = wn.
 Above, we showed that b^{wn} = x^{n}. By substitution, we have log_{b} x^{n} = wn.
 Above, we defined w: log_{b} x = w. By substitution, we have



 log_{b} x^{n} = (log_{b} x) n = nlog_{b} x.


We can use these properties to rewrite log expressions.
Expanding expressions
Using the properties we have derived above, we can write a log expression as the sum or difference of simpler expressions. Consider the following examples:
 log_{2} 8x = log_{2} 8 + log_{2} x = 3 + log_{2}x
 \begin{align*}log_3 \left (\frac{x^2} {3}\right ) = log_3 x^2  log_3 3 = 2log_3 x  1\end{align*}
Using the log properties in this way is often referred to as "expanding". In the first example, expanding the log allowed us to simplify, as log2 8 = 3. Similarly, in the second example, we simplified using the log properties, and the fact that log_{3} 3 = 1.
Example 1: Expand each expression:
a. \begin{align*}log_5 25x^2 y\end{align*}  b. \begin{align*}log_{10} \left (\frac{100x} {9b}\right )\end{align*} 

Solution:
 a. log_{5} 25x^{2} y = log_{5} 25 + log_{5}x^{2} + log_{5} y = 2 + 2 log_{5}x + log_{5} y

b.

\begin{align*}log_{10} \left (\frac{100x} {9b}\right )\end{align*} \begin{align*}= log_{10} 100x  log_{10} 9b\end{align*} \begin{align*}= log_{10} 100 + log_{10} x  \left [log_{10} 9 + log_{10} b\right]\end{align*} \begin{align*}= 2 + log_{10} x  log_{10} 9  log_{10} b\end{align*}

Just as we can write a single log expression as a sum and difference of expressions, we can also write expanded expressions as a single expression.
Condensing expressions
To condense a log expression, we will use the same properties we used to expand expressions. Consider the expression log_{6} 8 + log_{6} 27. Alone, each of these expressions does not have an integer value. The value of log_{6} 8 is between 1 and 2; the value of log_{6} 27 is also between 1 and 2. If we condense the expression, we get:

 log_{6} 8 + log_{6} 27 = log_{6} (8 × 27) = log_{6} 216 = 3
We can also condense algebraic expressions. This will be useful later for solving logarithmic equations.
Example 2: Condense each expression:

a. 2log_{3} x + log_{3} 5x  log_{3} (x + 1) b. log_{2} (x^{2}  4)  log_{2} (x + 2)
Solution:
a. \begin{align*}2log_3 x + log_3 5 x  log_3 (x + 1) = log_3 x_2 + log_3 5x  log_3 (x + 1)\end{align*}









 \begin{align*}= log_3 (x^2 (5x))  log_3 (x + 1)\end{align*}

















 \begin{align*}=log_3 \left (\frac{5x^3} {x + 1}\right )\end{align*}








b. \begin{align*}log_2 (x^2  4)  log_2 (x + 2)= log_2 \left (\frac{x^2  4} {x + 2}\right )\end{align*}








 \begin{align*}= log_2 \left (\frac{(x + 2)(x  2)} {x + 2}\right )\end{align*}















 \begin{align*}= log_2 (x  2)\end{align*}







It is important to keep in mind that a log expression may not be defined for certain values of x. First, the argument of the log must be positive. For example, the expressions in example 2b above are not defined for x ≤ 2 (which allows us to "cancel" (x+2) without worrying about the condition x≠ 2).
Second, the argument must be defined. For example, in example 2a above, the expression \begin{align*}\left (\frac{5x^3} {x + 1}\right )\end{align*} is undefined if x = 1.
The log properties apply to logs with any real base. Next we will examine logs with base 10 and base e, which are the most common bases for logs (though only one is actually called “common”).
Common logarithms and natural logs
A common logarithm is a log with base 10. We can evaluate a common log just as we evaluate any other log. A common log is usually written without a base.
Example 3: Evaluate each log
a. log 1  b. log 10  c. \begin{align*}log\sqrt{10}\end{align*} 

Solution:
 a. log 1 = 0, as 10^{0} = 1.
 b. log 10 = 1, as 10^{1} = 10
 c. \begin{align*}log \sqrt{10} = \frac{1} {2}\end{align*} because \begin{align*}\sqrt{10} = 10^{1/2}\end{align*}
As noted in lesson 3, logarithms were introduced in order to simplify calculations. After Napier introduced the logarithm, another mathematician, Henry Briggs, proposed that the base of a logarithm be standardized as 10. Just as Napier had labored to compile tables of log values (though his version of the logarithm is somewhat different from what we use today), Briggs was the first person to publish a table of common logs. This was in 1617!
Until recently, tables of common logs were included in the back of math textbooks. Publishers discontinued this practice when scientific calculators became readily available. A scientific calculator will calculate the value of a common log to 8 or 9 digits. Most calculators have a button that says LOG. For example, if you have TI graphing calculator, you can simply press LOG, and then a number, and the calculator will give you a log value up to 8 or 9 decimal places. For example, if you enter LOG(7), the calculator returns .84509804. This means that 10^{.84509804} ≈ 7. If we want to judge the reasonableness of this value, we need to think about powers of 10. Because 10^{1} = 10, log(7) should be less than 1.
Example 4: For each log value, determine two integers between which the log value should lie. Then use a calculator to find the value of the log.
a. log 50  b. log 818 

Solution:
 a. log 50

 The value of this log should be between 1 and 2, as 10^{1} = 10, and 10^{2} = 100.
 Using a calculator, you should find that log 50 ≈ 1.698970004.
 b. log 818

 The value of this log should be between 2 and 3, as 10^{2} = 100, and 10^{3} = 1000.

 Using a calculator, you should find that log 818 ≈ 2.912753304.
The calculator’s ability to produce log values is an example of the huge benefit that technology can provide. Only a few years ago, the calculations in the previous example would have each taken several minutes, while now they only take several seconds. While most people might not calculate log values in their every day lives, scientists and engineers are grateful to have such tools to make their work faster and more efficient.
Along with the LOG key on your calculator, you will find another logarithm key that says LN. This is the abbreviation for the natural log, the log with base e. Natural logs are written using “ln” instead of “log.” That is, we write the expression log_{e} x as In x. How you evaluate a natural log depends on the argument of the log. You can evaluate some natural log expressions without a calculator. For example, ln e = 1, as e^{1} = e. To evaluate other natural log expressions requires a calculator. Consider for example ln 7. Recall that e ≈ 2.7. This tells us that ln 7 should be slightly less than 2, as (2.7)^{2} = 7.29. Using a calculator, you should find that ln 7 ≈ 1.945910149.
Example 5: Find the value of each natural log.
a. \begin{align*}ln 100\end{align*}  b. \begin{align*}ln \sqrt{e}\end{align*} 

Solution:
 a. ln 100 is between 4 and 5. You can estimate this by considering powers of 2.7, or powers of 3: 3^{4} = 81, and 3^{5} = 243.

 Using a calculator, you should find that ln 100 ≈ 4.605171086.
 b. Recall that a square root is the same as an exponent of 1/2. Therefore \begin{align*}ln \sqrt{e} = ln (e^{1/2}) = 1/2\end{align*}
You may have noticed that the common log and the natural log are the only log buttons on your calculator. We can use either the common log or the natural log to find the values of logs with other bases.
Change of Base
Consider the log expression log_{3} 35. The value of this expression is approximately 3 because 3^{3} = 27. In order to find a more exact value of log_{3}35, we can rewrite this expression in terms of a common log or natural log. Then we can use a calculator.
Let’s consider a general log expression, log_{b} x = y. This means that b^{y} = x. Recall that if two expressions are equal, then the logs of the expressions are equal. We can use this fact, and the power property of logs, to write log_{b} x in terms of common logs.
\begin{align*}b^y = x \Rightarrow log b^y = log x\end{align*}  The logs of the expressions are equal 




\begin{align*}\Rightarrow y log b = log x\end{align*} Use the power property of logs \begin{align*}\Rightarrow y = \frac{log x} {log b}\end{align*} Divide both sides by \begin{align*}log b\end{align*} \begin{align*}\Rightarrow log_b x = \frac{log x} {log b}\end{align*} Substitute \begin{align*}log_b x = y\end{align*}


The final equation, \begin{align*}log_b x = \frac{log x} {log b}\end{align*}, is called the change of base formula. Notice that the proof did not rely on the fact that the base of the log is 10. We could have used a natural log. Thus another form of the change of base formula is \begin{align*}\mathit\log_b x = \frac{ln x} {ln b}\end{align*}.
Note that we could have used a log with any base, but we use the common log and the natural log so that we can use a calculator to find the value of an expression. Consider again log_{3} 35. If we use the change of base formula, and then a calculator, we find that

 \begin{align*}log_3 35 = \frac{log35} {log3} = 3.23621727\end{align*}.
Example 6: Estimate the value, and then use the change of base formula to find the value of log_{2} 17.
Solution: log_{2} 17 is close to 4 because 2^{4} = 16 and 2^{5} = 32. Using the change of base formula, we have \begin{align*}log_2 17 = \frac{log 17} {log 2}\end{align*}. Using a calculator, you should find that the approximate value of this expression is 4.087462841.
Lesson Summary
In this lesson we have developed and used properties of logarithms, including a formula that allows us to calculate the value of a log expression with any base. Out of context, it may seem difficult to understand the value of these kind of calculations. However, as you will see in later lessons in this chapter, we can use exponential and logarithmic functions to model a variety of phenomena.
Points to Consider
 Why is the common log called common? Why 10?
 Why would you want to estimate the value of a log before using a calculator to find its exact value?
 What kind of situations might be modeled with a logarithmic function?
Review Questions
 Expand the expression: log_{b} 5x^{2}
 Expand the expression: log_{3} 81x^{5}
 Condense the expression: log(x + 1) + log(x  1)
 Condense the expression: 3ln(x) + 2ln (y)  ln(5x  2)
 Evaluate the expressions:
 log 1000
 log 0.01
 Evaluate the expressions:
 ln e^{4}
 \begin{align*}ln \left (\frac{1} {e^9}\right )\end{align*}
 Use the change of base formula to find the value of log_{5} 100.
 What is the difference between log_{b} x^{n} and (log_{b}x)^{n}?
 Condense the expression in order to simplify: 3 log 2 + log 125
 Is this equation true for any values of x and y? log_{2} (x + y) = log_{2} x + log_{2} y If so, give the values. If not, explain why not.
Vocabulary
 Common logarithm
 A common logarithm is a log with base 10k. The log is usually written without the base.
 Natural logarithm
 A natural log is a log with base e. The natural log is written as ln.
 Scientific calculator
 A scientific calculator is an electronic, handheld calculator that will do calculations beyond the four operations (\begin{align*}+, , \times, \div \end{align*}), such as square roots and logarithms. Graphing calculators will do scientific operations, as well as graphing and equation solving operations.
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Date Created:
Feb 23, 2012Last Modified:
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