Question

Use the following steps to prove that ${\log }_{b}xy={\log }_{b}x+{\log }_{b}y$ a. Let $x=b^p$ and $y=b^q.$ Solve these expressions for $p$ and $q$ respectively. b. Use property E1 for exponents to express $xy$ in terms of $b,p$ and $q$ c. Compute ${\log }_{b}xy$ and simplify.

Short answer

Question: Prove that ${\log }_b(xy)={\log }_{b}x+{\log }_{b}y$ by following the given steps. Answer: The proof requires expressing x and y in terms of b, p, and q, expressing xy in terms of b, p, and q, and computing the log of xy and simplifying it. We obtain the following expression: ${\log }_b(xy)={\log }_{b}x+{\log }_{b}y$, which proves the logarithmic property.

Step-by-step solution

Express x and y in terms of b, p, and q

Given, $x=b^p$ and $y=b^q$. We need to find the expressions for p and q in terms of x and y. Using logarithm with base b, we have: $p={\log }_{b}x$ and $q={\log }_{b}y$

Express xy in terms of b, p, and q

Now we need to find the expression for xy in terms of b, p, and q. To do that, we can multiply x and y directly: $xy=b^p\cdot b^q$ Now, use property E1 for exponents, which states that $b^p\cdot b^q=b^{p+q}$: $xy=b^{p+q}$

Compute ${\log }_b(xy)$ and simplify

Now, we compute the logarithm of xy: ${\log }_b(xy)={\log }_b\left(b^{p+q}\right)$ Next, we use the property of logarithms which states that ${\log }_b\left(b^z\right)=z$ for any $z$, so we can simplify the expression: ${\log }_b(xy)=p+q$ Finally, we can substitute p and q from step 1: ${\log }_b(xy)={\log }_{b}x+{\log }_{b}y$ Hence proved, ${\log }_b(xy)={\log }_{b}x+{\log }_{b}y$.

Key concepts

Logarithmic Identities

Understanding logarithmic identities is fundamental to mastering how to work with logarithms. Identities allow us to simplify complex expressions and make calculations more manageable. One essential identity is the product rule for logarithms, which states that the logarithm of a product is the sum of the logarithms of the individual factors, expressed as ewline ewline ${\text{log}}_b(xy)={\text{log}}_b(x)+{\text{log}}_b(y)$.ewline ewline To prove this identity, we take advantage of the interplay between logarithms and exponents, a topic we delve into with more specifics in the upcoming sections. This particular identity simplifies the process of multiplying numbers in logarithmic form, which is otherwise a cumbersome task.

Exponent Properties

Exponents exhibit specific properties that greatly simplify calculations, especially when combined with logarithms. The product of powers property, for example, states that when you multiply powers with the same base, you can add the exponents, as shown by the equationewline ewline $b^p\cdot b^q=b^{p+q}$.ewline ewline This property is integral when we want to rewrite the multiplication of two exponential terms with the same base as a single exponential term. This is not only a crucial step in simplifying expressions but also forms the backbone of the proof in logarithmic identities, as we've seen in the example problem. Understanding this property highlights the deep connection between exponents and logarithms which are, in a sense, two sides of the same mathematical coin.

Logarithm Calculations

Calculating with logarithms can seem daunting, but it becomes much simpler when we apply the logarithmic identities we’ve learned. The process involves transforming the logarithmic expressions so that they utilize known identities to simplify the calculation. For instance, when you encounter the logarithm of a product, you can separate it into the sum of logarithms as established by the product rule. Similarly, when you find the logarithm of an exponent such as ${\text{log}}_b(b^{p+q})$, you can directly convert it to $p+q$ because of the inverse nature of logarithms and exponents. Such techniques turn log calculations from seemingly intricate puzzles into straightforward exercises.

Base and Exponent Relationship

The relationship between the base and the exponent in an exponential expression is the foundation for understanding logarithms. When we write an exponential equation, such as $x=b^p$, we show that the number $x$ is the result of scaling the base $b$ by the exponent $p$. In logarithmic terms, $p$ is the power we must raise the base $b$ to in order to achieve $x$, which is expressed by the inverse logarithm operation $p={\text{log}}_b(x)$. This direct relationship allows us to switch seamlessly between the exponential form and the logarithmic form, which is pivotal in solving equations that involve logarithms and in proving related mathematical identities.