Logarithmic Functions
In this article, we will discuss Logarithmic Functions. We will learn about:
- Converting between Exponential and Logarithmic form.
- Common and Natural Logarithms.
- Special Properties of Logarithms.
- Product and Quotient Laws of Logarithms.
This article is written to specifically meet the needs of students studying the Secondary 3 Additional Mathematics syllabus.
Investigation 1:
We know how to solve simple exponential equations.
Solve \(3^x =81\).
Solution:
To solve \(3^x =81\),
We can convert \(3^x =81\) as such:
\(3^x =3^4\)
The above equation has the same base, that is \(3\).
So, the value of \(x\) is \(4\) by simply doing a comparison.
But what about \(3^x = 26?\)
To solve \(3^x = 26\), we need to use logarithm as we cannot convert \(26\) into the power of \(3\).
Definition of Logarithms
The exponential form is expressed as \(a^b=c\) which is the form of
\(\begin{align} \text{base}^{\,power} = \text{term} \end{align}\).
In the logarithm form, \(a^b=c\) can be written as \(\begin{align} \log_{a}c=b \end{align}\) which is in the form of
\(\begin{align} \log_{\,base}{term} = power \end{align}\).
If you notice, in different forms, the base will switch from a base of an exponential term to a base of a logarithmic form on different sides of the equation and vice versa.
Suppose we have,
\(\begin{align} a^b=c \end{align}\)
So, how do we convert it to a logarithm?
\(\begin{align} \log_{a}c=b \end{align}\)
In the above logarithmic function, \(b\) and \(c\) swap places. \(\log\) indicates logarithmic function.
Example 1:
Convert \(\begin{align} 7^x = 12 \end{align}\) to logarithmic form.
Covert \(\begin{align} \log_8 x = 216 \end{align}\) to exponential form.
Solution:
- \(7^x = 12\).
In the above exponential form, the base is \(7\) and it will remain the same in the logarithmic form too. The power \(x\) and term \(12\), will flip their positions in a logarithmic form. We need to add \(\log\) that indicates a logarithmic form. So, the logarithmic form will be:
\(\log_7 12 = x\)
- \(\log_8 x = 216\)
To convert the above logarithmic form into exponential form, we need to identify the base. The base is \(8, 216\) is the power and \(x\) is the term. So, the exponential form is:
\(8^{216}=x\)
Question 1:
Convert \(\log_2 n = 7\) into exponential form.
Solution:
In \(\log_2 n = 7\), the base is \(2\). The \(n\) is the term and \(7\) is the power. So, the exponential form is:
\(2^7 = n\)
So, the correct answer is: \(2^7 = n\).
Question 2:
Convert \(y = 3^x\) into logarithmic form.
Solution:
In \(y = 3^x\) is written as \( 3^x=y\). The base is \(3\). The power is \(x\) and \(y\) is the term. We will add \(\log\) to indicate a logarithmic form. So, the logarithmic form is:
\(\log_3 y = x\)
So, the correct answer is: \(\log_3 y = x\)
Common Logarithm
Full |
Written |
---|---|
\(\log_{10}a\) |
\(\lg a\) |
\(\log_{10}7\) |
\(\lg 7\) |
In the above table, we use \(\lg\) to denote a common logarithm. The \(\lg\) denotes \(\log_{10}\).
Natural Logarithm
Full |
Written |
---|---|
\(\log_{e}a\) |
\(\ln a\) |
\(\log_{e}7\) |
\(\ln 7\) |
In the above table, we use \(\ln\) to denote a natural logarithm. In a natural logarithm, \(℮\) has a set value \((2.7182818…)\) and can be the base.
Example 2:
Use a calculator to evaluate the following expressions. Round off your answer to 3 significant figures.
- \(\lg 5\)
- \(\ln 16\)
Solution:
- \(\lg 5\) refers to \(\log_{10} 5\). Using the calculator:
\(\begin{align} \log_{10} 5 &= 0.69897...\\ &= 0.699 \text{ (3 s.f )} \end{align}\)
- \(\ln 16\) refers to \(\log_e 16\). Using the calculator:
\(\begin{align} \log_e 16 &= 2.77258...\\ &= 2.77 \text{ ( 3 s.f )} \end{align}\)
Question 3:
Use a calculator to evaluate the value of \(\begin{align} 4\lg10 - \frac {\ln2}{4} \end{align}\). Round off your answer to 3 significant figures.
Solution:
Using the calculator:
\(\begin{align} 4\lg10 - \frac {\ln2}{4} &= 4 - 0.17328...\\ &= 3.8267 \\ &= 3.83 \text{ ( 3.s.f )} \end{align}\)
So, the correct answer is \(3.83\).
Special properties of Logarithms
Index |
Logarithm |
---|---|
\(5^1 = 5\) |
\(\log_5 5 = 1\) |
\(a^1 = a\) |
\(\log_a a = 1\) |
\(5^0 = 5\) |
\(\log_5 1 = 0\) |
\(a^0 = a\) |
\(\log_a 1 = 0\) |
In the above table, the first two cases, \(\log_5 5 \) and \(\log_a a \) are both equal to 1. The next two cases, \(\log_5 1 \) and \(\log_a 1 \) are both equal to 0.
Example 3:
Without using a calculator, simplify \((\lg 10)^{12} \times 2\ln e -\ln 1\).
Solution:
Given:
\(\log_a a = 1\) and \(\log_a 1 = 0\).
\((\lg 10)^{12} \times 2\ln e -\ln 1\\ =(\log_{10}10)^{12}\times2\log_ee-\log_e1\\ =1^{12}\times 2(1)-0\\ =1^{12}\times 2(1)\\ =2\)
Question 4:
Simplify \((\log_7 1)^5 + \log_44 \times \ln 6\) without using the calculator.
Solution:
Given:
\(\log_a a = 1\) and \(\log_a 1 = 0\).
\((\log_7 1)^5 + \log_44 \times \ln 6\)
We can simply \((\log_7 1)^5\) using \(\log_a 1 = 0\).
We can simplify \((\log_44)\) using \(\log_a a = 1\)
\(=0^5 +1 \times \ln 6\)
\(=\ln 6\)
So, the correct answer is \(\ln 6\).
Investigation 2:
Calculate the value of the following expressions.
\(\lg 2 + \lg 3 = 0.77815...\) |
\(\lg 6 = 0.77815...\) |
\(\lg 3 + \lg 5 =1.17609...\) |
\(\lg 15 = 1.17609...\) |
\(\ln 3 + \ln 5 = 2.7080...\) |
\(\ln 15 = 2.7080...\) |
What do you notice?
From the above table, we can say that:
\(\log_a x + \log_a y = \log_a xy\).
Investigation 3:
\(\lg 6 - \lg 2 = 0.47712...\) |
\(\lg 3 = 0.47712...\) |
\(\lg 6 - \lg 3 = 0.3010...\) |
\(\lg 2 = 0.3010...\) |
\(\ln 6 - \ln 3 = 0.69314...\) |
\(\ln 2 = 0.69314...\) |
What do you notice?
From the above table, we can say that:
\(\log_a x - \log_a y = \log_a (\frac{x}{y})\).
Product and Quotient Laws of Logarithms
\(\log_a x + \log_a y = \log_a xy\).
This is the product law of logarithm.
\(\log_a x - \log_a y = \log_a (\frac{x}{y})\).
This is the quotient law of logarithm.
Note: The base should be the same for product and quotient laws of logarithms.
Example 4:
Without using the calculator, simplify the following expressions.
- \(\lg 7 + \lg 9\)
- \(\log_3 15 - \log_3 5\)
Solution:
- \(\lg 7 + \lg 9\)
Using the Product Law of Logarithm:
\(\begin{align} \log_a x + \log_a y &= \log_a xy \\ \lg 7 + \lg 9 &= \lg (7 \times 9 ) \\ &= \lg 63 \end{align}\).
- \(\log_3 15 - \log_3 5\)
Using the Quotient Law of Logarithm:
\(\begin{align} \log_a x - \log_a y &= \log_a \frac{x}{y} \\ \log_3 15 - \log_3 5 &= \log_3 \frac{15}{5} \\ &= \log_3 3\\ &= 1 \end{align}\).
Question 5:
Simplify \(\log_3 2 + \log _3 4 + \log_3 5\).
Solution:
Using the Product Law of Logarithm:
\(\log_a x + \log_a y = \log_a xy\).
So,
\(\begin{align} \log_3 2 + \log _3 4 + \log_3 5 &= \log_3 (2 \times 4 \times 5) \\ &= \log_3 40 \end{align}\)
So, the correct answer is \(\log_3 40\).
Conclusion
In the article, we learned about the conversion between exponential and logarithmic forms. We also understood the concept of common and natural logarithms and the special properties of logarithms. We studied the product and quotient laws of logarithms and solved various examples and questions to understand the concepts better.
Keep Learning! Keep Improving !!