WHAT IS A LOGARITHM?
Logarithm or log is another way of expressing exponents. A logarithm is an exponent (x) to which a base (b) must be raised to yield a given number (n). We can also say that logarithm is the inverse of exponentiation. When mathematically expressed, x is the logarithm of n to the base b if $b^{x}=n$, in which we can write as $\log _{b}n=x$.
The figure below shows the parts of a logarithmic and exponential equation with base b.
$\log _{b}n=x$ represents the logarithmic equation with base b and the form $b^{x}=n$ denotes the exponential equation with base b.
Say, for example, we have the exponential equation,
$3^{4}=81$
Since it is in the form $b^{x}=n$, the base is 3, the exponent is 4, and the argument is 81.
Using the same base, exponent, and argument, this exponential function is the same as:
$\log _{3}81=4$
wherein 3 denotes the base, 81 represents the argument, and 4 denotes the exponent.
The table below shows examples of equivalent logarithmic equation and exponential equation.
| Logarithmic Form | Exponential Form |
| $\log _{5}25=2$ | $5^{2}=25$ |
| $\log _{7}343=3$ | $7^{3}=343$ |
| $\log _{2}32=5$ | $2^{5}=32$ |
Note that when rewriting a logarithmic equation to an exponential equation or exponential equation to a logarithmic equation, always remember that the base of the logarithm is the same base of the exponent.
HISTORY OF LOGARITHM
In 1614, a Scottish baron named John Napier published a book titled Mirifici Logarithmorum Canonis Descriptio (Description of the Wonderful Rule of Logarithms), wherein the methods of logarithm were publicly introduced. In 1620, Joost Burgi, a Swiss craftsman, also published a book about logarithms. These two scholars, Napier and Joost, independently produced the methods of a logarithm to simplify mathematical calculations. Napier defined logarithms as a ratio of two distances in a geometric form, as opposed to the current definitions of logarithms as exponents. In 1685 and 1694, John Wallis and Johann Bernoulli were the ones who looked at the possibility of defining logarithms as exponents.
HOW DO WE EVALUATE LOGARITHMS?
To evaluate a logarithm of the form $\log _{b}(n)=x$ or $x=\log _{b}(n)$,
- Change the logarithmic form to the exponential form. Thus, $x=\log _{b}(n)\leftrightarrow b^{x}=n$
- Find the exponent x to which the base must be raised to get a value for n.
EXAMPLE #1
Evaluate $\log _{3}(243)=x$.
SOLUTION
Step 1: To find the value of x, change the logarithmic equation in the form $b^{x}=n$ where $b=3$ and $n=243$. Hence,
$3^{x}=243$
Step 2: Determine the exponent to which if we raise 3 to it, it will yield 243. Thus,
$3^{x}=3^{5}$
$x=5$
Therefore, $x=5$
EXAMPLE #2
What should be the value of x given the logarithmic equation $\log _{6}(1)=x$?
SOLUTION
Step 1: Change the given equation to exponential form where $b=6$ and $n=1$. Thus, we will get:
$6^{x}=1$
Step 2: Determine the value of x to which if we raise 6 to it, we will get 1. By rules of zero exponents, we know that any number raised to 0 will always be equal to 1. Therefore,
$6^{x}=6^{0}$
$x=0$
Therefore, $x=0$ .
EXAMPLE #3
What is the value of x in $x=\log _{4}(\frac{1}{16})$?
SOLUTION
Step 1: Change $x=\log _{4}(\frac{1}{16})$ to exponential form. Thus,
$4^{x}=\frac{1}{16}$
Step 2: We know that $4^{2}=16$, however, it is in the denominator. By negative exponent rule, we can say that:
$4^{x}=\frac{1}{4^{2}}$
$4^{x}=4^{-2}$
$x=-2$
Therefore, $x=-2$.
WHAT ARE THE RESTRICTIONS IN LOGARITHM?
The $\log _{b}n$ is defined when the base b is any positive number not equal to 1, and the argument n is positive. The following restrictions are made because of the rules between exponents and logarithms.
| Restrictions | Reason |
| $b>0$ | By definition, a logarithmic function must have a base b that is positive. |
| $a>0$ | Since a is the argument when b is raised to any positive number x, it makes sense that any positive number b raised to a power x that is also positive will yield a positive argument a. If $b^{x}>0$, it follows that $a>0$. |
| $b\neq 1$ | Suppose that b could be 1. Consider a logarithmic equation $\log _{1}(5)=x$. The equivalent exponential equation would be $1^{x}=5$ which will never be true since 1 raised to any power will always be 1. Thus, $b\neq 1$. |
WHAT IS THE IMPORTANCE OF LOGARITHM?
One of the most important factors of studying logarithm is its relationship to exponential functions. Logarithms can be used to solve exponential equations and functions. Some essential application of logarithm includes measuring the loudness in terms of decibels, measuring the intensity of earthquakes using the Richter scale, computing the brightness of stars, and computing for the pH balance and measure of acidity.
WHAT ARE THE PROPERTIES OF LOGARITHM?
There are three fundamental properties of logarithm, namely product rule, quotient rule, and power rule.
| Properties of Logarithm | |
| Product Rule | $\log {b}(mn)=\log {b}(m)+\log _{b}(n)$ |
| Quotient Rule | $\log {b}(\frac{m}{n})=\log {b}(m)-\log _{b}(n)$ |
| Power Rule | $\log {b}(m^{p})=p\times \log {b}(m)$ |
THE PRODUCT RULE
The product rule states that for any values of $m,n>0$ and $0<b\neq 1$,
$\log {b}(mn)=\log {b}(m)+\log _{b}(n)$.
This means that the logarithm of the product is the sum of the log of its factors.
Numerically, we can prove if this rule is plausible. Suppose $b=3$, $m=9$, and $n=27$. Then applying the product rule for logarithm will yield:
$\log {3}(9\times 27)=\log {3}(9)+\log _{3}(27)$
By substitution
$\log {3}(243)=\log {3}(9)+\log _{3}(27)$
Since $9\times 7=243$
$5=2+3$
By evaluating the three logarithms
$5=5$
Therefore, we can confirm that the product rule is plausible and true.
EXAMPLE #1
Prove that the product rule is plausible for the values of m, n, and b given $m=4$, $n=16$, and $b=2$.
SOLUTION
Step 1: Using the product rule, substitute the values of m, n, and b. Thus,
$\log {b}(mn)=\log {b}(m)+\log _{b}(n)$
$\log {2}(4\times 16)=\log {2}(4)+\log _{2}(16)$
$\log {2}(64)=\log {4}(m)+\log _{2}(16)$
Step 2: By evaluating $\log _{2}(64)$, $\log _{2}(4)$, and $\log _{2}(16)$,
6 = 2 + 4
6 = 6
Therefore, given the values of m, n, and b, the product rule is plausible.
EXAMPLE #2
What is the expanded form of $\log _{7}(4y)$?
SOLUTION
Using the product rule, note that $b=7$ and the argument is a product of 4 and y. Hence, we can say that $m=4$ and $n=y$. Thus, expanding $\log _{7}(4y)$ will give us:
Therefore, the expanded form of $\log _{7}(4y)$ is $\log {7}(4y)=\log {7}(4)+\log _{7}(y)$.
EXAMPLE #3
Determine the simplified form of $\log {5}(8)+\log {5}(y)$.
SOLUTION
By product rule for logarithm, given $\log {5}(8)+\log {5}(y)$ where $b=5$, $m=8$, and $n=y$, the compressed form is $\log _{5}(8y)$.
Therefore, simplifying $\log {5}(8)+\log {5}(y)$ will give us $\log _{5}(8y)$.
Note that before compressing logarithm expressions using the product rule, the bases of all the logarithms must always be the same. Say for example the expression $\log {7}(21)+\log {8}(3)$. This logarithmic expression cannot be simplified using product rule since they have a different base.
THE QUOTIENT RULE
The quotient rule states that for any values of $m,n>0$ and $0<b\neq 1$,
$\log {b}(\frac{m}{n})=\log {b}(m)-\log _{b}(n)$.
This implies that the logarithm of a quotient is equal to the difference between the logarithm of the numerator and denominator.
Suppose $m=216$, $n=36$, and $b=6$, then by quotient rule,
$\log {6}(\frac{216}{36})=\log {6}(216)-\log _{6}(36)$
By substitution
$\log {6}(6)=\log {6}(216)-\log _{6}(36)$
Since $\frac{216}{36}$
$1=3-2$
By evaluating the three logarithms
$1=1$
Therefore, we can say that the quotient rule is plausible and true.
EXAMPLE #1
Prove that the quotient rule is true such that $m=64$, $n=8$, and $b=2$
SOLUTION
Step 1: By quotient rule, substitute the values of m, n, and b. Hence,
$\log {b}(\frac{m}{n})=\log {b}(m)-\log _{b}(n)$
$\log {2}(\frac{64}{8})=\log {2}(64)-\log _{2}(8)$
$\log {2}(8)=\log {2}(64)-\log _{2}(8)$
Step 2: Evaluate $\log _{2}(8)$, $\log _{2}(64)$, and $\log _{2}(8)$,
$3=6-3$
$3=3$
Therefore, given the values of m, n, and b, the product rule is plausible.
EXAMPLE #2
What is the expanded form of $$?
SOLUTION
Given $b=9$, $m=81$, and $n=y$, then by quotient rule, the expanded form is the difference between two logarithms. Thus,
$\log {9}(\frac{81}{y})=\log {9}(81)-\log _{9}(y)$
Therefore, the expanded form of $\log _{9}(\frac{81}{y})$ is $\log {9}(\frac{81}{y})=\log {9}(81)-\log _{9}(y)$.
EXAMPLE #3
Simplify $\log {7}(y)-\log {7}(3)$.
SOLUTION
By quotient rule, given $\log {7}(y)-\log {7}(3)$ where $b=7$, $m=y$, and $n=3$, the compressed form is $\log _{7}(\frac{y}{3})$.
Therefore, the simplified form of $\log {7}(y)-\log {7}(3)$ is $\log _{7}(\frac{y}{3})$.
Like the restriction in the product rule, always make sure that before compressing logarithmic expressions by quotient rule, the bases of all logarithms must be the same.
THE POWER RULE
The power rule states that for any values of $m,n>0$ and $0<b\neq 1$,
$\log {b}(m^{p})=p\times \log {b}m$.
This means that the logarithm of a power is equal to the exponent multiplied by the logarithm of the base of the power.
Suppose $b=5$, $m=5$, and $p=2$, then applying the power rule,
$\log {5}(5^{2})=2\times \log {5}5$
By substitution
$\log {5}(25)=2\times \log {5}5$
Since $5^{2}=25$
$2=2\times1$
By evaluating three logarithms
$2=2$
Therefore, we can confirm that the power rule is plausible and true.
EXAMPLE #1
Confirm that the power rule is plausible given $b=2$, $m=4$, and $p=3$.
SOLUTION
Step 1: Using the power rule, substitute the values of b, m, and p. Thus,
$\log {b}(m^{p})=p\times \log {b}m$
$log {2}(4^{3})=3\times \log {2}(4)$
$\log {2}(64)=3\times \log {2}(4)$
Step 2: Evaluate $\log _{2}(64)$ and $\log _{2}(4)$.
$6=3\times 2$
$6=6$
Therefore, given the values of b, m, and p, the power rule is plausible.
EXAMPLE #2
Expand the logarithmic expression $\log _{8}(y^{4})$
SOLUTION
Given $b=9$, $n=y$, and $m=81$, then by power rule, the expanded form of $\log _{8}(y^{4})$ is
$\log {8}(y^{4})=4\times \log {8}(y)$
$\log {8}(y^{4})=4 \log {8}(y)$
Therefore, expanding $\log _{8}(y^{4})$ will result to $\log {8}(y^{4})=4\log {8}(y)$.
EXAMPLE #3
What is the simplified form of the logarithmic expression $3\log _{8}(4)$?
SOLUTION
Using the power rule, given $3\log _{8}(4)$ where $b=8$, $m=4$, and $p=3$, the compressed form is $\log _{8}(4^{3})$.
Therefore, the simplified form of $3\log _{8}(4)$ is $\log _{8}(4^{3})$ or $\log _{8}(64)$.
EXPANDING LOGARITHMIC EXPRESSIONS
When it comes to expanding logarithmic expressions with multiple properties, the first thing to do is work out all possible properties that can be done from the inner parts to the outer part of the expression.
For example, we have a logarithmic expression $\log _{3}(\frac{4y^{2}}{9})$. How can we expand this?
The first thing we need to do is use the quotient rule to separate $4y^{2}$ and 9. Hence,
$\log {3}(\frac{4y^{2}}{9})=\log {3}(4y^{2})-\log _{3}(9)$.
Then, the expression $\log _{3}(4y^{2})$ has an argument that is a product of 4 and $y^{2}$. Thus, by product rule, $\log {3}(4y^{2})=\log {3}(4)+\log _{3}(y^{2})$. Therefore, we now have
$\log {3}(\frac{4y^{2}}{9})=\log {3}(4)+\log {3}(y^{2})-\log {3}(9)$.
Lastly, in the expression $\log _{3}(y^{2})$ we still have an argument that is raised in the power of 2. Therefore, we can use the power rule to expand this expression. By power rule, $\log {3}(y^{2})=2\log {3}(y)$ which will result to
$\log {3}(\frac{4y^{2}}{9})=\log {3}(4)+2\log {3}(y)-\log {3}(9)$.
Therefore, the expanded form of $\log {3}(\frac{4y^{2}}{9})$ is $\log {3}(4)+2\log {3}(y)-\log {3}(9)$
EXAMPLE
Expand the logarithmic expression $\log _{x}(\frac{y^{3}}{2z})$.
SOLUTION
Step 1: Use the quotient rule.
$\log {x}(\frac{y^{3}}{2z})=\log {x}(y^{3})-\log _{x}(2z)$
Step 2: By power rule, we can expand $\log _{x}(y^{3})$ to $3\log _{x}(y)$. Hence,
$\log {x}(\frac{y^{3}}{2z})=3\log {x}(y)-\log _{x}(2z)$
Step 3: The expression $\log _{x}(2z)$ can still be expanded using the product rule. Thus, $\log {x}(2z)=\log {x}(2)+\log _{x}(z)$. Therefore,
$\log {x}(\frac{y^{3}}{2z})=3\log {x}(y)-[\log {x}(2)+\log {x}(z)]$
The reason why there is a grouping symbol [ ] for $\log {x}(2)+\log {x}(z)$ for the reason that the product rule only applies to the denominator, and between the argument 2z. Hence, the grouping.
Therefore, the expanded form of $\log _{x}(\frac{y^{3}}{2z})$ is $3\log {x}(y)-[\log {x}(2)+\log _{x}(z)]$.
SIMPLIFYING LOGARITHMIC EXPRESSIONS
To simplify logarithmic expressions, you must always check the bases of the given logarithmic expressions. Then, work out the logarithmic expressions one at a time. Take note of grouping symbols such as ( ), [ ], and { } as it will make a huge difference when simplifying logarithmic expressions. Lastly, always remember that + means the use of product rule, and – means the use of quotient rule.
Suppose we have the logarithmic expression $4\log {5}(y)-3\log {5}(8)$, how are we going to compress this?
Since both logarithmic expressions are in base 5, we can proceed to the next step. In both expressions $4\log _{5}(y)$ and $3\log _{5}(8)$, we can use the power rule. Thus,
$4\log {5}(y)=\log {5}(y^{4})$ and $3\log {5}(8)=\log {5}(8^{3})$.
The operator between them is – which means it satisfies the quotient rule. Therefore, combining $4\log _{5}(y)$ and $3\log _{5}(8)$ using the quotient rule will yield $\log _{5}(\frac{y^{4}}{8^{3}})$.
Therefore, the simplified form of $4\log {5}(y)-3\log {5}(8)$ is $\log _{5}(\frac{y^{4}}{8^{3}})$ or $\log _{5}(\frac{y^{4}}{512})$.
EXAMPLE
Simplify the logarithmic expression $5\log {9}(x)+3\log {9}(y)$.
SOLUTION
Step 1: Check if both logarithmic expressions have the same base. Since both of them are in base 9, then it is possible to compress.
Step 2: Use the power rule for $5\log _{9}(x)$ and $4\log _{9}(y)$. Thus,
$5\log {9}(x)=\log {9}(x^{5})$ and $4\log {9}(y)=\log {5}(y^{4})$.
Step 3: The operator that exists between the original logarithmic expression is +. Thus, we will use the product rule to simplify. To do this, multiply the arguments that exist between $\log _{9}(x^{5})$ and $\log _{5}(y^{4})$. Multiplying the arguments will yield to $x^{5}\times y^{4}= x^{5}y^{4}$. Hence,
$5\log {9}(x)+4\log {9}(y)=\log _{9}(x^{5}y^{4})$
Therefore, the simplified form of the logarithmic expression $5\log {9}(x)+4\log {9}(y)$ is $\log _{9}(x^{5}y^{4})$ wherein $(x^{5}y^{4})$ is the argument.
WHAT ARE SPECIAL LOGARITHMS?
There are two types of special logarithms namely, common logarithm and natural logarithm. While we know that the base of a logarithm can be any positive number, special logarithms have bases that are often used than others.
COMMON LOGARITHM
Logarithms with base 10 are called common logarithms. When mathematically written, we usually omit the base since it is already understood that it is in base 10. For example, given $\log_{10}(n)$ we can already write it as $\log _{n}$.
NATURAL LOGARITHM
The natural logarithm is a logarithm whose base is the Euler’s number (e). Natural logarithms can be written as either $\log_{e}(n)$ or $\ln (e)$.
The table below summarizes the things you should know about common and natural logarithms.
| Base | Regular Notation | Special Notation | |
| Common Logarithm | 10 | $\log_{10}(n)$ | $\log _{n}$ |
| Natural Logarithm | e | $\log_{e}(n)$ | $\ln (e)$ |
The properties of logarithm also apply to special logarithms.
WHAT ARE LOGARITHMIC IDENTITIES?
The table below shows some of the logarithmic identities derived from the three fundamental properties of logarithms.
| Identity | Reason |
| $\log _{b}(1)=0$ | This identity came from the fact that any number raised to 0 will always be 1 or also known as the rule of zero exponents. |
| $\log _{b}(b)=1$ | The reason behind this identity is the fact that any base number raised to 1 will always be equal to itself. |
| $\log _{b}(b^{p})=p$ | |
| $b^{\log _{b}(n)}=n$ |
WHAT IS THE CHANGE OF BASE FORMULA?
The logarithmic change of base formula is used to write a logarithm of a number wherein the argument is not a rational power of the base. The change-of-base formula is denoted by:
$\log {b}(a)=\frac{\log {c}(a)}{\log _{c}(b)}$
wherein
a in the argument of the logarithm is the same as the a in the argument of the numerator;
b is the base of the original logarithm is the same as the b in the argument of the numerator;
c is the base of both logarithms in the numerator and denominator wherein it can be any positive number not equal to 1.
The easiest way to use the change base formula is to change the logarithms to base 10.
EXAMPLE #1
Evaluate the value of $\log _{16}(4)$.
SOLUTION
Step 1: Using the change-of-base formula by changing the base of the logarithm to base 10. Note that $\log_{10}$ is the same as log. Thus,
$\log _{16}(4)=\frac{\log (4)}{\log (16)}$
Step 2: The denominator $\log (16)$ can be expressed as $\log (4^{2})$ since $16=4^{2}$. Thus,
$\log _{16}(4)=\frac{\log (4)}{\log (4^{2})}$
Step 3: By power rule, we can expand the logarithm in the denominator as $4^{2}=2\log (4)$. Hence,
$\log _{16}(4)=\frac{\log (4)}{2\log (4))}$
Step 4: Cancel $\log (4)$ from the numerator and denominator. Thus,
$\log _{16}(4)=\frac{1}{2}$
Therefore, the value of $\log _{16}(4)$ is $\frac{1}{2}$.
EXAMPLE #2
Estimate the value of $\log _{7}(21)$.
SOLUTION
Step 1: Use the change of base formula and change the logarithm to base 10. Thus,
$\log _{7}(21)=\frac{\log (21)}{\log (7)}$
Step 2: Since 21 is not a rational power of 21, we can use the calculator to compute for the value of $\frac{\log (21)}{\log (7)}$. Hence,
$\frac{\log (21)}{\log (7)}=1.564$
Therefore, the estimated value of $\log _{7}(21)$ is 1.56.
Recommended Worksheets
Working with Radicals and Integer Exponents 8th Grade Math Worksheets
Understanding Exponents 6th Grade Math Worksheets
Multiplication of Functions (Travel and Tours Themed) Worksheets
