Primer on Logarithms

Logarithms are the "inverse of exponentiation." If you’ve ever wondered “what power of a number produces another number?”, you’re working with logarithms. This primer introduces logarithms, explains their properties, and proves those properties step by step.

What is a Logarithm?

The logarithm answers the question:
"To what power must the base be raised to produce a given number?"

Mathematically:

$$\log_b(x) = y \quad \text{means} \quad b^y = x$$

• $b$: the base of the logarithm (e.g., 2, 10, or $e$).
• $x$: the number you’re taking the logarithm of.
• $y$: the exponent.

Example:

$\log_2(8) = 3$ because $2^3 = 8$.

Key Properties of Logarithms

Logarithms simplify complex operations. Below are the fundamental properties, along with proofs.

1. Product Rule

$$\log_b(xy) = \log_b(x) + \log_b(y)$$

Proof:

1. Let $\log_b(x) = p$ and $\log_b(y) = q$. By the definition of logarithms: $$b^p = x \quad \text{and} \quad b^q = y$$
2. The product $xy$ is: $$xy = b^p \cdot b^q$$
3. By the laws of exponents, $b^p \cdot b^q = b^{p+q}$, so: $$xy = b^{p+q}$$
4. Take $\log_b$ of both sides: $$\log_b(xy) = \log_b(b^{p+q})$$
5. By the property $\log_b(b^k) = k$, we get: $$\log_b(xy) = p + q$$
6. Substitute back $p = \log_b(x)$ and $q = \log_b(y)$: $$\log_b(xy) = \log_b(x) + \log_b(y)$$

2. Quotient Rule

$$\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y)$$

Proof:

1. Let $\log_b(x) = p$ and $\log_b(y) = q$. By the definition of logarithms: $$b^p = x \quad \text{and} \quad b^q = y$$
2. The quotient $\frac{x}{y}$ is: $$\frac{x}{y} = \frac{b^p}{b^q}$$
3. By the laws of exponents, $\frac{b^p}{b^q} = b^{p-q}$, so: $$\frac{x}{y} = b^{p-q}$$
4. Take $\log_b$ of both sides: $$\log_b\left(\frac{x}{y}\right) = \log_b(b^{p-q})$$
5. By the property $\log_b(b^k) = k$, we get: $$\log_b\left(\frac{x}{y}\right) = p - q$$
6. Substitute back $p = \log_b(x)$ and $q = \log_b(y)$: $$\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y)$$

3. Power Rule

$$\log_b(x^k) = k \cdot \log_b(x)$$

Proof:

1. Let $\log_b(x) = p$. By the definition of logarithms: $$b^p = x$$
2. Raise both sides to the power $k$: $$(b^p)^k = x^k$$
3. By the laws of exponents, $(b^p)^k = b^{pk}$, so: $$x^k = b^{pk}$$
4. Take $\log_b$ of both sides: $$\log_b(x^k) = \log_b(b^{pk})$$
5. By the property $\log_b(b^k) = k$, we get: $$\log_b(x^k) = pk$$
6. Substitute back $p = \log_b(x)$: $$\log_b(x^k) = k \cdot \log_b(x)$$

4. Change of Base Formula

$$\log_b(x) = \frac{\log_k(x)}{\log_k(b)}$$

Proof:

1. Let $\log_b(x) = p$. By the definition of logarithms: $$b^p = x$$
2. Take $\log_k$ of both sides (using a new base $k$): $$\log_k(b^p) = \log_k(x)$$
3. By the power rule, $\log_k(b^p) = p \cdot \log_k(b)$: $$p \cdot \log_k(b) = \log_k(x)$$
4. Solve for $p$: $$p = \frac{\log_k(x)}{\log_k(b)}$$
5. Recall that $p = \log_b(x)$, so: $$\log_b(x) = \frac{\log_k(x)}{\log_k(b)}$$

Applications of Logarithms

Logarithms have widespread applications:

• Growth and Decay: Population dynamics, radioactive decay, and compound interest.
• Scales: Decibels (sound), pH (acidity), and earthquake magnitudes.
• Computing: Algorithm efficiency (e.g., $O(\log n)$), binary trees, and compression.

Quick Practice Problems

1. Prove $\log_b(1) = 0$.
2. Simplify $\log_2(16)$.
3. Solve for $x$: $10^{\log(x)} = 1000$.