Logarithms Properties & Laws: Concepts
Definitions & Assumptions
Recall that if $a > 0$, $a \neq 1$, and $x > 0$, then: $$\log_a(x) = y \iff a^y = x$$
To ensure all proofs and identities are mathematically defined, we assume:
- The base $a > 0$ and $a \neq 1$.
- The numbers inside logarithms, $M$ and $N$, are positive real numbers ($M > 0$, $N > 0$).
Properties & Proofs
1. Product Rule
Converting to exponential form:
$a^x = MN$, $a^y = M$, and $a^z = N$.
Substitute $M$ and $N$ into the product:
$MN = a^y \cdot a^z = a^{y+z}$.
Since $a^x = MN$, we get $a^x = a^{y+z} \Rightarrow x = y + z$.
Replacing back the values:
$\log_a(MN) = \log_a(M) + \log_a(N)$.
2. Quotient Rule
Converting to exponential form:
$a^y = M$ and $a^z = N$.
Divide the equations:
$\frac{M}{N} = \frac{a^y}{a^z} = a^{y-z}$.
Taking log on both sides:
$\log_a\left(\frac{M}{N}\right) = y - z = \log_a(M) - \log_a(N)$.
3. Power Rule
$\Rightarrow a^x = M^k$ and $a^y = M$.
Substitute $M = a^y$ into $a^x = M^k$:
$a^x = (a^y)^k = a^{ky} \Rightarrow x = ky$.
Replacing values:
$\log_a(M^k) = k \log_a(M)$.
4. Change of Base Formula
Take log to base $b$ on both sides:
$\log_b(a^x) = \log_b(M) \Rightarrow x \log_b(a) = \log_b(M)$.
Solve for $x$:
$x = \frac{\log_b(M)}{\log_b(a)} \Rightarrow$ $\log_a(M) = \frac{\log_b(M)}{\log_b(a)}$.
5. Log of 1
Since $a^0 = 1$ for any positive $a \neq 1$, we get $x = 0$.
Therefore, $\log_a(1) = 0$.
6. Log of a number to the same base
Since $a^1 = a$, we get $x = 1$.
Therefore, $\log_a(a) = 1$.
Summary of Logarithm Properties
| Property Name | Logarithmic Notation | What it does |
|---|---|---|
| Product Rule | $\log_a(MN) = \log_a(M) + \log_a(N)$ | Turns multiplication into addition |
| Quotient Rule | $\log_a(M/N) = \log_a(M) - \log_a(N)$ | Turns division into subtraction |
| Power Rule | $\log_a(M^k) = k \log_a(M)$ | Exponents come down |
| Log of same base | $\log_a(a) = 1$ | Log of base itself is 1 |
| Log of 1 | $\log_a(1) = 0$ | Logarithm of 1 to any base is 0 |
| Base Changing | $\log_a(n) = \frac{\log_b(n)}{\log_b(a)}$ | Changes base from $a$ to $b$ |
- $\log_a(m+n) \neq \log_a(m) + \log_a(n)$ (Logs do NOT distribute over addition!)
- $\log_a(m-n) \neq \log_a(m) - \log_a(n)$ (Logs do NOT distribute over subtraction!)
- If $x = y$, then $\log_a(x) = \log_a(y)$.
(a) $\log_7(3) + \log_7(5)$
(b) $\log_2(9) - \log_2(3)$
(c) $\log_4(3) + \log_4(6) - 3\log_4(2)$
(d) $\log_3(15) - \log_3(5)$
Show Solution
(b) $\log_2(9) - \log_2(3) = \log_2\left(\frac{9}{3}\right) = \mathbf{\log_2(3)}$
(c) $\log_4(3) + \log_4(6) - \log_4(2^3) = \log_4(18) - \log_4(8) = \log_4\left(\frac{18}{8}\right) = \mathbf{\log_4\left(\frac{9}{4}\right)}$
(d) $\log_3(15) - \log_3(5) = \log_3\left(\frac{15}{5}\right) = \mathbf{\log_3(3)} = \mathbf{1}$
(a) $\log_7(343)$ (b) $\log_3(27\sqrt{3})$
Show Solution
(b) $27\sqrt{3} = 3^3 \cdot 3^{1/2} = 3^{3 + 1/2} = 3^{7/2}$.
$\Rightarrow \log_3(3^{7/2}) = \frac{7}{2}\log_3(3) = \mathbf{\frac{7}{2}}$
Show Solution
Method 2: $\log_3\left(\frac{81}{9}\right) = \log_3(9) = \log_3(3^2) = \mathbf{2}$.
Common Logarithms (Base 10)
Logarithms with base 10 are called common logarithms. They are widely used in scientific scales, such as the Richter scale (earthquakes), pH scale (acidity), and decibel scale (sound loudness).
By convention, $\log_{10}(x)$ is written as simply $\log(x)$. When no base is written, base 10 is assumed.
Rules of Common Logarithms:
- $\log(xy) = \log(x) + \log(y)$
- $\log(x/y) = \log(x) - \log(y)$
- $\log(x^n) = n\log(x)$
- $\log(1) = 0$
- $\log(10) = 1$
(a) $\log(2) + \log(7)$ (b) $\log(6) - \log(3)$
Show Solution
(b) $\log(6) - \log(3) = \log\left(\frac{6}{3}\right) = \mathbf{\log(2)}$
(a) $\log_3\left(\frac{4}{7}\right)$ (b) $\log_3(28)$ (c) $\log_3\left(\frac{7}{3}\right)$
Show Solution
(b) $\log_3(28) = \log_3(4 \times 7) = \log_3(4) + \log_3(7) = \mathbf{b + a}$
(c) $\log_3\left(\frac{7}{3}\right) = \log_3(7) - \log_3(3) = a - 1 = \mathbf{a - 1}$
(a) $3\log(5) - 2$ (b) $1 + \log(2)$
Show Solution
Substitute $2 = 2\log(10) = \log(10^2) = \log(100)$.
$\Rightarrow \log(5^3) - \log(100) = \log(125) - \log(100) = \log\left(\frac{125}{100}\right) = \mathbf{\log\left(\frac{5}{4}\right)}$
*(Note: If the base is 2 instead of 10, i.e., $3\log_2(5) - 2 = \log_2(125) - \log_2(4) = \log_2(125/4)$). Let's follow the standard base-10 calculation.
(b) $1 + \log(2)$:
Substitute $1 = \log(10)$.
$\Rightarrow \log(10) + \log(2) = \log(10 \times 2) = \mathbf{\log(20)}$
- Common logarithms scale numbers that are very small or very large. For example, if quantities range from $0.0000000001$ to $10,000,000,000$, their common logarithms compress this huge range down to just $[-10, 10]$.
- Natural Logarithms use base $e$ ($e \approx 2.71828$, an irrational mathematical constant). They are denoted by $\ln(x)$, i.e., $\log_e(x) = \ln(x)$, and are fundamental in calculus.