O Level (Sec 3 & 4)Additional Mathematics
Logarithms
Master logarithms with this comprehensive study guide. Learn the laws of logarithms, change of base formula, and solve logarithmic equations. Includes 8 worked examples with step-by-step video solutions and real-world applications.
By Timothy Gan24 November 2021Singapore O-Level syllabus
Free revision pack
Use the notes online, then download the worksheet to practise without distractions.
- 7 concept sections
- 8 worked questions
- 14 video lessons
7
Concept sections
8
Practice questions
6
Concept videos
8
Solution videos
Start here
Understanding Logarithms
Hi students, let's explore the concept of logarithms today! A logarithm is the inverse function to an exponential function. It is defined as the exponent to which a fixed base must be raised to yield a given number.
Logarithms are applied in many real-life problem-solving situations, from measuring earthquake magnitudes on the Richter scale to calculating sound intensity in decibels, and measuring acidity and alkalinity with pH values.
In this comprehensive guide, I will explain the fundamental laws of logarithms, demonstrate conversions between logarithmic and exponential forms, and provide you with plenty of practice questions.
Logarithms is a very common topic that can be found in many mathematics syllabi such as the IGCSE Additional Mathematics (0606) and Singapore SEAB Additional Mathematics.
Video lesson
Study Guide - Understanding Logarithms and Conversions
1
Concept
Conversion Between Logarithmic and Exponential Form
An exponential equation is just a special case of a logarithmic equation. Understanding the relationship between these two forms is fundamental to working with logarithms.
The conversion formula is:
$a^b = c \Leftrightarrow b = \log_a(c)$
This means that if $a$ raised to the power of $b$ equals $c$, then $b$ is the logarithm of $c$ to the base $a$.
Special cases to remember:
- Common logarithm: $\log_{10}(x) = \lg x$ (base of 10)
- Natural logarithm: $\log_e(x) = \ln x$ (base $e \approx 2.718$)
For example:
- $2^3 = 8$ can be written as $\log_2(8) = 3$
- $10^2 = 100$ can be written as $\lg 100 = 2$
- $e^1 = e$ can be written as $\ln e = 1$
2
Concept
Special Properties of Logarithms
There are two important special properties that you should memorize when working with logarithms:
- Logarithm of 1: $\log_a(1) = 0$ for any base $a$
- Logarithm of the Base: $\log_a(a) = 1$ for any base $a$
These properties are very useful for simplifying logarithmic expressions and solving equations.
Video lesson
Study Guide - Special Properties of Logarithms
3
Concept
Product Law of Logarithms
The Product Law states that the sum of the logarithms of two numbers is equal to the logarithm of the product of those two numbers.
Formula: $\log_a(b) + \log_a(c) = \log_a(b \times c)$
Proof:
Let $\log_a(b) = x$ and $\log_a(c) = y$
Let $\log_a(b) = x$ and $\log_a(c) = y$
Converting to exponential form:
$a^x = b$ and $a^y = c$
$a^x = b$ and $a^y = c$
Multiplying both equations:
$a^x \times a^y = b \times c$
$a^x \times a^y = b \times c$
Using the law of indices:
$a^{x+y} = b \times c$
$a^{x+y} = b \times c$
Converting back to logarithmic form:
$x + y = \log_a(b \times c)$
$x + y = \log_a(b \times c)$
Substituting back:
$\log_a(b) + \log_a(c) = \log_a(b \times c)$
$\log_a(b) + \log_a(c) = \log_a(b \times c)$
Example: $\log_2(3) + \log_2(4) = \log_2(3 \times 4) = \log_2(12)$
Video lesson
Study Guide - Product Law and Its Proof
4
Concept
Quotient Law of Logarithms
The Quotient Law states that the difference between the logarithms of two numbers is equal to the logarithm of the quotient of those two numbers.
Formula: $\log_a(b) - \log_a(c) = \log_a\left(\frac{b}{c}\right)$
Proof:
Let $\log_a(b) = x$ and $\log_a(c) = y$
Let $\log_a(b) = x$ and $\log_a(c) = y$
Converting to exponential form:
$a^x = b$ and $a^y = c$
$a^x = b$ and $a^y = c$
Dividing both equations:
$\frac{a^x}{a^y} = \frac{b}{c}$
$\frac{a^x}{a^y} = \frac{b}{c}$
Using the law of indices:
$a^{x-y} = \frac{b}{c}$
$a^{x-y} = \frac{b}{c}$
Converting back to logarithmic form:
$x - y = \log_a\left(\frac{b}{c}\right)$
$x - y = \log_a\left(\frac{b}{c}\right)$
Substituting back:
$\log_a(b) - \log_a(c) = \log_a\left(\frac{b}{c}\right)$
$\log_a(b) - \log_a(c) = \log_a\left(\frac{b}{c}\right)$
Example: $\log_5(25) - \log_5(5) = \log_5\left(\frac{25}{5}\right) = \log_5(5) = 1$
Video lesson
Study Guide - Quotient Law and Its Proof
5
Concept
Power Law of Logarithms
The Power Law allows us to move the exponent to the front of the logarithm as a coefficient.
Formula: $\log_a(b^r) = r \cdot \log_a(b)$
Proof:
Let $\log_a(b) = x$
Let $\log_a(b) = x$
Converting to exponential form:
$a^x = b$
$a^x = b$
Raising both sides to the power of $r$:
$(a^x)^r = b^r$
$(a^x)^r = b^r$
Using the law of indices:
$a^{rx} = b^r$
$a^{rx} = b^r$
Converting back to logarithmic form:
$rx = \log_a(b^r)$
$rx = \log_a(b^r)$
Substituting back:
$r \cdot \log_a(b) = \log_a(b^r)$
$r \cdot \log_a(b) = \log_a(b^r)$
Example: $\log_2(8^3) = 3 \cdot \log_2(8) = 3 \cdot 3 = 9$
This law is particularly useful when dealing with square roots and fractional powers:
$\log_a(\sqrt{b}) = \log_a(b^{1/2}) = \frac{1}{2}\log_a(b)$
$\log_a(\sqrt{b}) = \log_a(b^{1/2}) = \frac{1}{2}\log_a(b)$
Video lesson
Study Guide - Power Law and Its Proof
6
Concept
Change of Base Formula
The change-of-base formula allows us to convert any logarithm into another logarithm with a different base. This is especially useful when you need to use a calculator that only has $\log$ (base 10) and $\ln$ (base $e$) buttons.
Formula: $\log_a(b) = \frac{\log_c(b)}{\log_c(a)}$
where $c$ is any positive base you choose (typically 10 or $e$).
Proof:
Let $\log_a(b) = x$
Let $\log_a(b) = x$
Converting to exponential form:
$a^x = b$
$a^x = b$
Taking logarithm of both sides with base $c$:
$\log_c(a^x) = \log_c(b)$
$\log_c(a^x) = \log_c(b)$
Using the Power Law:
$x \cdot \log_c(a) = \log_c(b)$
$x \cdot \log_c(a) = \log_c(b)$
Solving for $x$:
$x = \frac{\log_c(b)}{\log_c(a)}$
$x = \frac{\log_c(b)}{\log_c(a)}$
Substituting back:
$\log_a(b) = \frac{\log_c(b)}{\log_c(a)}$
$\log_a(b) = \frac{\log_c(b)}{\log_c(a)}$
Special Property: $\log_x(y) = \frac{1}{\log_y(x)}$
This can be derived by setting $c = y$ in the change of base formula:
$\log_x(y) = \frac{\log_y(y)}{\log_y(x)} = \frac{1}{\log_y(x)}$
$\log_x(y) = \frac{\log_y(y)}{\log_y(x)} = \frac{1}{\log_y(x)}$
Example: To find $\log_2(10)$ using a calculator:
$\log_2(10) = \frac{\lg 10}{\lg 2} = \frac{1}{0.301} \approx 3.32$
$\log_2(10) = \frac{\lg 10}{\lg 2} = \frac{1}{0.301} \approx 3.32$
Video lesson
Study Guide - Change of Base and Its Proof
7
Concept
Real-World Applications of Logarithms
The logarithm is a very common and useful way to measure the relative magnitudes of things, especially when dealing with quantities that vary over a very wide range.
The Richter Scale: A measure of the energy released in an earthquake. Each whole number increase on the Richter scale represents a tenfold increase in measured amplitude and approximately 31.6 times more energy release.
Decibel (dB): A logarithmic measure of the intensity of a sound. The decibel scale is logarithmic because the human ear responds logarithmically to sound intensity.
pH Value: A measure of acidity and alkalinity on a scale of 0 to 14. pH is defined as $pH = -\log_{10}[H^+]$, where $[H^+]$ is the hydrogen ion concentration.
Half-Life in Radioactive Decay: The time it takes for half of a radioactive substance to decay can be calculated using logarithms.
These applications demonstrate why logarithms are so important in science, engineering, and everyday life.
Guided practice
Practice Questions with Video Solutions
Attempt each question before opening the solution. Then compare your method with the worked steps and video explanation.
Question 1Video solution
Conversion to Logarithmic Form
Question
Convert each of the following to logarithmic form.
(a) $x^{2y} = 4$
(b) $5^x = 12$
(c) $(4x)^{5-p} = a$
(a) $x^{2y} = 4$
(b) $5^x = 12$
(c) $(4x)^{5-p} = a$
Video lesson
Study Guide - Logarithms Question 1
Step-by-step solution
- 1Recall the conversion formula: $a^b = c \Leftrightarrow b = \log_a(c)$
- 2(a) From $x^{2y} = 4$, we identify: base = $x$, exponent = $2y$, result = $4$
- 3Converting to logarithmic form: $2y = \log_x(4)$
- 4(b) From $5^x = 12$, we identify: base = $5$, exponent = $x$, result = $12$
- 5Converting to logarithmic form: $x = \log_5(12)$
- 6(c) From $(4x)^{5-p} = a$, we identify: base = $4x$, exponent = $5-p$, result = $a$
- 7Converting to logarithmic form: $5-p = \log_{4x}(a)$
Final answer
(a) $2y = \log_x(4)$, (b) $x = \log_5(12)$, (c) $5-p = \log_{4x}(a)$
Question 2Video solution
Conversion to Exponential Form and Solving
Question
Solve each of the following equations.
(a) $\ln 4x = 5$
(b) $\log_x(16) = 4$
(c) $\ln(3x) = 6$
(d) $\log_k(81) = 2$
(a) $\ln 4x = 5$
(b) $\log_x(16) = 4$
(c) $\ln(3x) = 6$
(d) $\log_k(81) = 2$
Video lesson
Study Guide - Logarithms Question 2
Step-by-step solution
- 1(a) From $\ln 4x = 5$, convert to exponential form: $e^5 = 4x$
- 2Solving for $x$: $x = \frac{e^5}{4} \approx 37.1$
- 3(b) From $\log_x(16) = 4$, convert to exponential form: $x^4 = 16$
- 4Taking fourth root: $x = \sqrt[4]{16} = 2$ (since $2^4 = 16$)
- 5(c) From $\ln(3x) = 6$, convert to exponential form: $e^6 = 3x$
- 6Solving for $x$: $x = \frac{e^6}{3} \approx 134.7$
- 7(d) From $\log_k(81) = 2$, convert to exponential form: $k^2 = 81$
- 8Taking square root: $k = 9$ (positive value only)
Final answer
(a) $x = \frac{e^5}{4}$, (b) $x = 2$, (c) $x = \frac{e^6}{3}$, (d) $k = 9$
Question 3Video solution
Simplifying Using Product Law
Question
Simplify $\log_6(2) + \log_6(3)$ if possible.
Video lesson
Study Guide - Logarithms Question 3
Step-by-step solution
- 1Apply the Product Law: $\log_a(b) + \log_a(c) = \log_a(b \times c)$
- 2Both logarithms have the same base (6), so we can apply the law.
- 3$\log_6(2) + \log_6(3) = \log_6(2 \times 3)$
- 4$= \log_6(6)$
- 5Using the special property: $\log_a(a) = 1$
- 6$= 1$
Final answer
$1$
Question 4Video solution
Simplifying Using Quotient Law
Question
Simplify $\log_5(25) - \log_5(5)$ if possible.
Video lesson
Study Guide - Logarithms Question 4
Step-by-step solution
- 1Apply the Quotient Law: $\log_a(b) - \log_a(c) = \log_a\left(\frac{b}{c}\right)$
- 2Both logarithms have the same base (5), so we can apply the law.
- 3$\log_5(25) - \log_5(5) = \log_5\left(\frac{25}{5}\right)$
- 4$= \log_5(5)$
- 5Using the special property: $\log_a(a) = 1$
- 6$= 1$
Final answer
$1$
Question 5Video solution
Evaluating with Power Law
Question
Evaluate each of the following without using the calculator.
(a) $\log_8(64)$
(b) $\log_2(\sqrt{4}) + \log_2(\sqrt{3}) - \log_2(\sqrt{6})$
(a) $\log_8(64)$
(b) $\log_2(\sqrt{4}) + \log_2(\sqrt{3}) - \log_2(\sqrt{6})$
Video lesson
Study Guide - Logarithms Question 5
Step-by-step solution
- 1(a) We need to find: $\log_8(64)$
- 2Since $8 = 2^3$ and $64 = 2^6$, we can write: $64 = 8^2$
- 3Therefore: $\log_8(64) = \log_8(8^2) = 2$
- 4(b) First, convert square roots to fractional powers:
- 5$\log_2(\sqrt{4}) + \log_2(\sqrt{3}) - \log_2(\sqrt{6})$
- 6$= \log_2(4^{1/2}) + \log_2(3^{1/2}) - \log_2(6^{1/2})$
- 7Using Power Law: $\log_a(b^r) = r \cdot \log_a(b)$
- 8$= \frac{1}{2}\log_2(4) + \frac{1}{2}\log_2(3) - \frac{1}{2}\log_2(6)$
- 9$= \frac{1}{2}[\log_2(4) + \log_2(3) - \log_2(6)]$
- 10Using Product and Quotient Laws:
- 11$= \frac{1}{2}\log_2\left(\frac{4 \times 3}{6}\right) = \frac{1}{2}\log_2(2) = \frac{1}{2} \times 1 = \frac{1}{2}$
Final answer
(a) $2$, (b) $\frac{1}{2}$
Question 6Video solution
Change of Base Formula
Question
(a) Given that $p = \log_a(9)$, find $\log_3(a)$ in terms of $p$
(b) If $p = \lg 14$, find $\log_{14}\left(1\frac{2}{5}\right)$ in terms of $p$
(b) If $p = \lg 14$, find $\log_{14}\left(1\frac{2}{5}\right)$ in terms of $p$
Video lesson
Study Guide - Logarithms Question 6
Step-by-step solution
- 1(a) Given: $p = \log_a(9)$, find $\log_3(a)$
- 2Using the change of base formula: $\log_a(9) = \frac{\log_3(9)}{\log_3(a)}$
- 3Since $9 = 3^2$, we have $\log_3(9) = 2$
- 4Therefore: $p = \frac{2}{\log_3(a)}$
- 5Rearranging: $\log_3(a) = \frac{2}{p}$
- 6(b) Given: $p = \lg 14$, find $\log_{14}\left(1\frac{2}{5}\right) = \log_{14}\left(\frac{7}{5}\right)$
- 7Using the change of base formula:
- 8$\log_{14}\left(\frac{7}{5}\right) = \frac{\lg\left(\frac{7}{5}\right)}{\lg 14}$
- 9$= \frac{\lg 7 - \lg 5}{p}$
- 10Since $14 = 2 \times 7$: $p = \lg 14 = \lg 2 + \lg 7$
- 11Therefore: $\lg 7 = p - \lg 2$
- 12$= \frac{p - \lg 2 - \lg 5}{p} = \frac{p - \lg 10}{p} = \frac{p - 1}{p}$
Final answer
(a) $\log_3(a) = \frac{2}{p}$, (b) $\log_{14}\left(1\frac{2}{5}\right) = \frac{p-1}{p}$
Question 7Video solution
Solving Simultaneous Logarithm Equations
Question
Solve the simultaneous equations:
$\log_4(x) - \log_2(y) = 2$
$3^x = 81\left(9^{\frac{3}{2}-3y}\right)$
$\log_4(x) - \log_2(y) = 2$
$3^x = 81\left(9^{\frac{3}{2}-3y}\right)$
Video lesson
Study Guide - Logarithms Question 7
Step-by-step solution
- 1From equation 1: $\log_4(x) - \log_2(y) = 2$
- 2Convert $\log_4(x)$ to base 2: $\log_4(x) = \frac{\log_2(x)}{\log_2(4)} = \frac{\log_2(x)}{2}$
- 3Substituting: $\frac{\log_2(x)}{2} - \log_2(y) = 2$
- 4Multiply by 2: $\log_2(x) - 2\log_2(y) = 4$
- 5$\log_2(x) - \log_2(y^2) = 4$
- 6$\log_2\left(\frac{x}{y^2}\right) = 4$, so $\frac{x}{y^2} = 2^4 = 16$ ... (i)
- 7From equation 2: $3^x = 81\left(9^{\frac{3}{2}-3y}\right)$
- 8Since $81 = 3^4$ and $9 = 3^2$:
- 9$3^x = 3^4 \times 3^{2(\frac{3}{2}-3y)}$
- 10$3^x = 3^4 \times 3^{3-6y}$
- 11$3^x = 3^{7-6y}$
- 12Therefore: $x = 7 - 6y$ ... (ii)
- 13Substitute (ii) into (i): $\frac{7-6y}{y^2} = 16$
- 14$7 - 6y = 16y^2$
- 15$16y^2 + 6y - 7 = 0$
- 16Factoring: $(2y-1)(8y+7) = 0$
- 17$y = \frac{1}{2}$ or $y = -\frac{7}{8}$ (reject negative)
- 18From (ii): $x = 7 - 6(\frac{1}{2}) = 7 - 3 = 4$
Final answer
$x = 4$, $y = \frac{1}{2}$
Question 8Video solution
Applications - Radioactive Decay
Question
The mass, $m$ grams, of a radioactive substance, present at time $t$ days after first being observed, is given by the formula $m = 28e^{-0.00072t}$.
(a) Find the value of $m$ when $t = 20$
(b) Find the value of $t$ when the mass is half of its value at $t = 0$
(c) State the value which $m$ approaches as $t$ becomes very large
(d) Sketch the graph of $m$ against $t$
(a) Find the value of $m$ when $t = 20$
(b) Find the value of $t$ when the mass is half of its value at $t = 0$
(c) State the value which $m$ approaches as $t$ becomes very large
(d) Sketch the graph of $m$ against $t$
Video lesson
Study Guide - Logarithms Question 8
Step-by-step solution
- 1(a) When $t = 20$:
- 2$m = 28e^{-0.00072(20)} = 28e^{-0.0144} \approx 28(0.9857) \approx 27.6$ grams
- 3(b) At $t = 0$: $m = 28e^0 = 28$ grams
- 4Half of this is $14$ grams. When $m = 14$:
- 5$14 = 28e^{-0.00072t}$
- 6$\frac{14}{28} = e^{-0.00072t}$
- 7$0.5 = e^{-0.00072t}$
- 8Taking natural logarithm of both sides:
- 9$\ln(0.5) = -0.00072t$
- 10$t = \frac{\ln(0.5)}{-0.00072} = \frac{-0.693}{-0.00072} \approx 963$ days
- 11(c) As $t \to \infty$, $e^{-0.00072t} \to 0$
- 12Therefore, $m \to 0$ as $t$ becomes very large
- 13(d) The graph starts at $(0, 28)$ and decreases exponentially, approaching $m = 0$ asymptotically as $t$ increases. This is a typical exponential decay curve.
Final answer
(a) $m \approx 27.6$ grams, (b) $t \approx 963$ days, (c) $m \to 0$, (d) Exponential decay curve from $(0,28)$ approaching the $t$-axis
Revision summary
Key Formulas to Remember
Conversion Formula
$a^b = c \Leftrightarrow b = \log_a(c)$
Fundamental relationship between exponential and logarithmic forms
Product Law
$\log_a(b) + \log_a(c) = \log_a(b \times c)$
Sum of logs equals log of product
Quotient Law
$\log_a(b) - \log_a(c) = \log_a\left(\frac{b}{c}\right)$
Difference of logs equals log of quotient
Power Law
$\log_a(b^r) = r \cdot \log_a(b)$
Exponent can be moved to the front as a coefficient
Change of Base Formula
$\log_a(b) = \frac{\log_c(b)}{\log_c(a)}$
Convert to any base $c$ (typically 10 or $e$)
Identities and Results to Memorise
- $\log_a(1) = 0$
- $\log_a(a) = 1$
- $\ln(1) = 0$
- $\ln(e) = 1$
- $\log_x(y) = \frac{1}{\log_y(x)}$
- $\lg x = \log_{10}(x)$
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