Laws of logarithms are rules that can be applied to simplify and solve complicated logarithmic equations. When manipulating logarithms, it is important to make sure all the bases are the same.
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
1. 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. Thus, using our exponential (indices) rule of \(M^n \cdot M^n = M^{m+n}\),
\(ab = (x^c)(x^d) = x^{c+d}\)
\(ab = x^{c+d}\)
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 (\(\log_x(x^{c+d})\)), 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
1. 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 \(\frac{M^m}{M^n} = M^{m-n}\),
\(\frac{a}{b} = \frac{x^c}{x^d} = x^{c-d}\)
\(\frac{a}{b} = x^{c-d}\)
3. Take the log of both sides.
\(\log_x(\frac{a}{b}) = \log_x(x^{c-d})\)
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 .
\(\log_x(x^{c-d}) = c-d\)
\(\log_x(\frac{a}{b}) = c-d\)
5. We defined c and d in part 1 where \(\log_x(a) = c\) and \(\log_x(b) = d\):
\(c-d = \log_x(a) - \log_x(b)\)
\(\log_x(\frac{a}{b}) = \log_x(a) - \log_x(b)\)
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
\(\log_b(a^k) = k\log_b(a)\) which 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 equationk\(\log_a(x)\)
\(\log_a(x) = \frac{\log_b(x)}{\log_b(a)}\)
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, \(x = 1 \pm i\sqrt{8}\)
Laws of Logs - Key takeaways
- The four main laws you need to be familiar with are product law, quotient law, change of base law, and power rule.
- Reciprocal law, log of a base and log of 1 are logarithms that are more specialised – they can only be used in specific contexts.
Explanations 
Exams 
Magazine 
