The laws of logs offer a powerful toolset for mathematicians, scientists, and problemsolvers of all backgrounds. Through the understanding and application of these laws, we can simplify complex logarithmic expressions, manipulate equations, and solve intricate mathematical problems.
The versatility of logarithms extends into various fields, including physics, engineering, finance, and computer science, enabling us to reason with exponential growth, assess resource depletion, model financial trends, and analyze algorithms.
The basic laws of logs
 Product (addition) Law: \(\log_a(m) + \log_b(n) = \log_a(mn)\)
 Quotient Law: \(\log_a(m)  \log_b(n) = \log_a(\frac{m}{ n})\)
 Power Law: \(\log_a(x^b) = b\log_a(x)\)
 Change of Law: \(\log_a(x) = \frac{\log_b(x)}{\log_b(a)}\)
The change of law formula is in the formula booklet you are given in the exam.
Other logarithms
 Reciprocal Laws: \(\log_a(\frac{1}{x}) = \log(x^{1}) = \log(x)\)
 Log of the base: \(\log_a(a)=1\)
 Log of 1: \(\log_a(1) = 0\)

Although technically a logarithm law, it is important to remember logarithms can be converted into exponentials: \(\log_a(b) = x\) can be written as \(a^x = b\).
Law of logs proof
It is not necessary to be able to prove each logarithm law for the exam, but it is important to understand each step and why it occurs.
Product (addition) law
If \(log_x(a) = c\) and \(log_x(b) = d\), then you can rewrite the logarithms as an exponential function.
For \(log_x(a) = c\), the base is x, the exponent is c, the answer to the exponential is a.
Therefore, it can be written as \(x^c = a\)
For \(\log_x(b) = d\), the base is x, the exponent is d, and the answer to the exponential is b.
Therefore, it can be written as \(x^d = b\)
2. Using our exponential rules, we have "=m".
\(ab = (x^c)(x^d) = x^{c+d}\)
\(ab = x^{c+d}\)
3. Take the log of both sides:
\(\log_x(ab) = \log_x(x^{c+d})\)
4. Because \(\log_x(x^{c+d})\) includes both an exponential with a base of x and a logarithm with a base of x, they will cancel each other out to become just c + d.
\(log_x(x^{c+d}) = c + d \)
\(\log_x(ab) = c + d\)
This step is because logarithms and exponentials are inverse functions of each other. Think about when we cancel out the +4 and 4 in x +4 4 = 10 – this is the same principle.
5. We defined \(c\) and \(d\) in part 1. \(\log_x(a) = c\) and \(\log_x(b) = d\)
Therefore, \(c + d = \log_x(a) + \log_x(b)\)
\(\log_x(ab) = \log_x(a) + \log_x(b)\)
Quotient rule
If \(\log_x(a) = c\) and \(\log_x(b) = d\), then you can rewrite the logarithms as an exponential function.
For \( \log_x(a) = c \), the base is x, the exponent is c, and the answer to the exponential is a. Therefore, it can be written as \( x^c = a \).
For \( \log_x(b) = d \), the base is x, the exponent is d, and the answer to the exponential is b.
Therefore, it can be written as \(x^d = b\)
2. Thus, using our exponential (indices) rules of \( \frac{M^m}{M^n} = M^{mn} \),
\( \frac{a}{b} = \frac{x^c}{x^d} = x^{cd} \)
\( \frac{a}{b} = x^{cd} \)
3. Take the log of both sides.
\( \log_x(\frac{a}{b}) = \log_x(x^{cd}) \)
4. Because \(\log_x(x^{cd})\) includes both an exponential with a base of x and a logarithm with a base of x, they will cancel each other out to become just c  d.
\(\log_x(x^{cd}) = cd\)
\(\log_x(\frac{a}{b}) = cd\)
5. We defined c and d in part 1 where \(\log_x(a) = c\) and \(\log_x(b) = d\):
cd = \(\log_x(a)  \log_x(b)\)
\(\log_x(\frac{a}{b}) = \log_x(a)  \log_x(b)\)
Change of base
1. Let \(log_a(x) = k\) where the base is a, the exponent is k, and the answer to the exponential = x.
Therefore, it can be rewritten as an exponential: \(a^k = x\)
2. Take the log of both sides \(log_b(a^k) = log_b(x)\)
3. Use the power rule to simplify
When \( \log_b(a^k) = k\log_b(a) \), you can then substitute back into the equation
\( k\log_b(a) = \log_b(x) \)
4. Rearrange to get k on its own by dividing through \( k \log_b(a) \)
\( k = \frac{\log_b(x)}{\log_b(a)} \)
5. As k is already defined, it can be substituted into the equation
\( k\log_a(x) \)
\( \log_a(x) = \frac{\log_b(x)}{\log_b(a)} \)
Reciprocal law
 \( \log_a(\frac{1}{x}) \) can be written as \( \log_a(x^{1}) \) using our exponential rules with negatives.
 You can use the power log rule to bring the 1 down so \( \log_a(x^{1}) \) becomes \( \log_a(x) \).
Log of the base
 Set \(\log_a(a)=x\) where the base is a, the exponent is x, and the answer to the exponential is a. Therefore, it can be written as \(a^x = a\).
 According to exponential rules, if the answer of an exponential is equal to the base, then the exponent must be 1.
Log of 1
 Set \( \log_a1=x \) where the base is a, the exponent is x, and the answer to the exponential is 1. Therefore, it can be written as \(a^x = 1\).
 According to exponential rules, if the answer of an exponential is 1, then the exponent must be 0.
Simplifying and solving using laws of logs
Here, we will go through some examples of simplifying a range of laws of logs.
Simplifying and solving using 1 log law
Show \(\log (6) + \log (4) = \log (24)\)
\(\log (6) + \log (4) = \log (6 \cdot 4) = \log (24)\)
Solve \(\log (14)  \log (7)\)
\(\log (14)  \log (7) = \log(\frac{14}{7}) = \log (2) = 0.301 (3 s.f)
Simplify \(2\log (9)\), keep in exact form
\(2\log(9) = \log(9)^2 = \log(81)\)
Solve \(2\log(2\cdot 3)\)
\(2\log (2 \cdot 3) = \log(2 \cdot 3)^2 = \log(6)^2 = \log(36) = 1.56 (3 s.f)
Simplifying and solving using multiple log laws
It might help to use rules that simplify individual logs before doing the simplifying multiple log laws.
Solve \(3\log(4)  \log(8)\)
\(\log(4)^3 = \log(64)\log(64)  \log(8) = \log(\frac{64}{8}) = \log(8) = 0.903 (3 s.f)\)
Simplify \(\log_4(4x^2)\)
\(\log_4(4) + \log_4(x^2)1+ 2\log_4x\)
Prove \(x = 1 \pm i\sqrt{8}\) where \(2\log_2(x+3)  \log_2(x) = 3\)
1. Using the power rule, \(2\log_2(x+3) = \log_2(x+3)^2\)
Therefore, \(\log_2(x+3)^2  \log_2(x) = 3\)
2. Using the quotient rule, \(\frac{\log_2(x+3)^2}{\log_2(x)} = 3\)
3. When you want to remove the logarithm, you need to convert it into an exponential. This works in the same way as normal – just make sure you label each part.
Base is 2; exponent is 3; answer to exponential is \(\frac{(x+3)^2}{x}\)
\(2^3 = \frac{(x+3)^2}{x}\)
4. Solve like a normal equation
\(8 = \frac{(x+3)^2}{x}\)
\(8x = (x+3)^2\)
\(0 = x^2 2x+9\)
Using the formula we get, \(x = 1 \pm i\sqrt{8}\)