Question

Use the following steps to prove that \(\log _{b}(x y)=\log _{b} x+\log _{b} y\). a. Let \(x=b^{p}\) and \(y=b^{q}\). Solve these expressions for \(p\) and \(q\) respectively. b. Use property El for exponents to express \(x y\) in terms of \(b, p\) and \(q\). c. Compute \(\log _{b}(x y)\) and simplify.

Short answer

Question: Prove that \(\log_b(xy) = \log_b(x) + \log_b(y)\) where x and y are positive real numbers, and b is a positive real number such that \(b \neq 1\). Answer: \(\log_b(xy) = \log_b(x) + \log_b(y)\)

Step-by-step solution

Express x and y in terms of p and q

We are given that \(x=b^p\) and \(y=b^q\). We are asked to solve these expressions for p and q respectively. By definition of logarithms, we can write: $$ p = \log_b(x) $$ and $$ q = \log_b(y) $$

Use property E1 for exponents to express xy in terms of b, p, and q

Now, we need to express \(xy\) in terms of \(b, p,\) and \(q\). Using the property E1 for exponents, we have: $$ x \cdot y = b^p \cdot b^q $$ Using the laws of exponents (specifically, that \(b^p \cdot b^q = b^{p+q}\)), we can rewrite the expression as: $$ x \cdot y = b^{p + q} $$

Compute the logarithm base b of xy and simplify

Next, we need to compute \(\log_b(xy)\) using the expression for xy we derived in step 2: $$ \log_b(xy) = \log_b(b^{p+q}) $$ By the property of logarithm, we can bring down the exponent \(p+q\) to the front: $$ \log_b(xy) = (p+q) \log_b(b) $$ Since \(\log_b(b) = 1\), the expression simplifies to: $$ \log_b(xy) = p+q $$ And finally, substituting \(p = \log_b(x)\) and \(q = \log_b(y)\) from step 1, we have: $$ \log_b(xy) = \log_b(x) + \log_b(y) $$ Hence, we have proven that \(\log_b(xy) = \log_b(x) + \log_b(y)\).

Key concepts

Laws of exponents

When dealing with powers and exponents, the laws of exponents are essential building blocks. These rules help us simplify expressions and solve complex mathematical problems. One of the most important laws to understand is the product of powers rule. This rule states that when you multiply two exponents with the same base, you can add their exponents. So, if you have \( b^p \cdot b^q \), you can rewrite it as \( b^{p+q} \). This is because multiplication involves combining like terms, and you're essentially adding the number of times the base is used as a factor for each exponent.

Other helpful laws of exponents include:

• Quotients of powers: \( \frac{b^p}{b^q} = b^{p-q} \)
• Power of a power: \( (b^p)^q = b^{pq} \)
• Power of a product: \( (xy)^p = x^p \cdot y^p \)

Understanding these laws makes manipulating and solving for variables in exponential expressions much more manageable. By applying these rules, as shown in our exercise, we expressed \( xy \) in terms of \( b^{p+q} \), simplifying the problem significantly.

Logarithmic identities

Logarithms are the inversion of exponential functions and they have specific identities that make solving logarithmic problems straightforward. One highly useful identity is the product property of logarithms. It states that the logarithm of a product is the sum of the logarithms of its factors: \( \log_b(xy) = \log_b(x) + \log_b(y) \). This identity simplifies multiplication inside the log into addition outside the log.

Logarithms also have other identities which include:

• Quotient property: \( \log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y) \)
• Power property: \( \log_b(x^y) = y \cdot \log_b(x) \)
• Change of base formula: \( \log_b(x) = \frac{\log_k(x)}{\log_k(b)} \)

These identities are not just mathematical curiosities—they provide us powerful tools for solving real-world problems involving exponential growth, interest rates, sound intensity, and more. In the exercise, by using these properties, we saw how \( \log_b(xy) \) was simplified to \( \log_b(x) + \log_b(y) \), showcasing how the product property works.

Mathematical proof

Mathematical proofs are logical arguments that confirm the truth of a statement through a sequence of valid steps or deductions. A proof is essentially a way to demonstrate that a certain property or equation holds in all specified cases. There are various forms of proofs, such as direct, indirect, contradiction, and mathematical induction, but they all share a common goal: to conclusively establish truth.

The proof in our exercise showcases a direct proof, where the original statement \(\log_b(xy) = \log_b(x) + \log_b(y)\) was shown to be true by manipulating expressions using property rules of exponents and logarithms.

• Start by expressing terms using exponential forms (like \( x = b^p \), \( y = b^q \))
• Apply exponent rules to find \( xy \) in simpler terms (\( b^{p+q} \))
• Compute the log using derived expressions and adjust using property of logs (\( \log_b(b^{p+q}) \))
• Simplify to conclusion (match with \( p+q \), refuting with original log terms)

Developing a correct proof means ensuring every step is backed by a theorem or acceptable math principle. This helps build a strong foundation for understanding more sophisticated mathematical concepts and encourages logical problem-solving skills.