Deduce the following laws of logarithms from the laws of exponents: a. logb(xy) = logb x logb(y) b. logb (x/y) = logb(x)/ logb(y) c. logb (x^y) = ylogb(x)

We can deduce the following logarithmic laws from the corresponding laws of exponents by leveraging the inverse relationship between exponents and logarithms. Let’s derive each rule one by one.

### a. **Product Rule**:  
\[
\log_b(xy) = \log_b(x) + \log_b(y)
\]

#### Proof:
Start by assuming:
\[
\log_b(x) = p \quad \text{and} \quad \log_b(y) = q
\]
This implies that:
\[
b^p = x \quad \text{and} \quad b^q = y
\]
Now, using the property of exponents that \( b^p \cdot b^q = b^{p+q} \), we get:
\[
xy = b^p \cdot b^q = b^{p+q}
\]
Taking the logarithm of both sides (base \( b \)):
\[
\log_b(xy) = \log_b(b^{p+q})
\]
By the definition of a logarithm, \( \log_b(b^z) = z \), so:
\[
\log_b(xy) = p + q
\]
Substituting \( p = \log_b(x) \) and \( q = \log_b(y) \), we get:
\[
\log_b(xy) = \log_b(x) + \log_b(y)
\]
This proves the product rule of logarithms.

---

### b. **Quotient Rule**:  
\[
\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y)
\]

#### Proof:
Start by assuming:
\[
\log_b(x) = p \quad \text{and} \quad \log_b(y) = q
\]
This implies:
\[
b^p = x \quad \text{and} \quad b^q = y
\]
Now, using the property of exponents \( \frac{b^p}{b^q} = b^{p-q} \), we get:
\[
\frac{x}{y} = \frac{b^p}{b^q} = b^{p-q}
\]
Taking the logarithm of both sides (base \( b \)):
\[
\log_b\left(\frac{x}{y}\right) = \log_b(b^{p-q})
\]
By the definition of the logarithm, \( \log_b(b^z) = z \), so:
\[
\log_b\left(\frac{x}{y}\right) = p - q
\]
Substituting \( p = \log_b(x) \) and \( q = \log_b(y) \), we get:
\[
\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y)
\]
This proves the quotient rule of logarithms.

---

### c. **Power Rule**:  
\[
\log_b(x^y) = y \log_b(x)
\]

#### Proof:
Start by assuming:
\[
\log_b(x) = p
\]
This implies:
\[
b^p = x
\]
Now, using the property of exponents \( (b^p)^y = b^{py} \), we get:
\[
x^y = (b^p)^y = b^{py}
\]
Taking the logarithm of both sides (base \( b \)):
\[
\log_b(x^y) = \log_b(b^{py})
\]
By the definition of the logarithm, \( \log_b(b^z) = z \), so:
\[
\log_b(x^y) = py
\]
Substituting \( p = \log_b(x) \), we get:
\[
\log_b(x^y) = y \log_b(x)
\]
This proves the power rule of logarithms.

---

Would you like further details or have any questions?

Here are some related questions you may want to explore:
1. How does the change of base formula for logarithms relate to the laws of exponents?
2. How can logarithmic identities help simplify complex expressions?
3. How does the natural logarithm (base \(e\)) fit within these rules?
4. What are some real-life applications of logarithmic laws?
5. How are logarithmic and exponential functions used in solving equations?

**Tip:** Remember, logarithms and exponents are inverse operations, which makes it easier to translate rules from one to the other!

Understand how to deduce logarithmic laws such as the product, quotient, and power rules from the corresponding laws of exponents. This problem explores core algebraic concepts, including logarithms and their relationship with exponents. Suitable for students in Grades 9-12.

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