Problem 28 - Precalculus Mathematics for Calculus

Question

Use the Laws of Logarithms to expand the expression. $$\log _{2}(x y)^{10}$$

Step-by-Step Solution

Answer

\( 10 \log_{2}(x) + 10 \log_{2}(y) \)

Step 1: Apply the Power Rule of Logarithms

The first step is to apply the power rule of logarithms, which states \( \log_b(M^n) = n \log_b(M) \). So, we can bring down the exponent 10 as a coefficient:\[ \log_{2}((xy)^{10}) = 10 \log_{2}(xy) \].

Step 2: Apply the Product Rule of Logarithms

Now, let's use the product rule of logarithms, which states \( \log_b(MN) = \log_b(M) + \log_b(N) \). Here, \( M = x \) and \( N = y \), so we can expand it as:\[ 10 \log_{2}(xy) = 10 (\log_{2}(x) + \log_{2}(y)) \].

Step 3: Distribute the Coefficient

Finally, distribute the coefficient 10 to both terms inside the parentheses:\[ 10 (\log_{2}(x) + \log_{2}(y)) = 10 \log_{2}(x) + 10 \log_{2}(y) \].

Key Concepts

Power Rule of Logarithms

The power rule of logarithms is a powerful tool that simplifies the management of exponents when dealing with logarithms. It essentially allows us to "bring down" any exponent present in a logarithmic argument. This rule is mathematically represented by the formula:

• \( \log_b(M^n) = n \log_b(M) \)

What this formula tells us is that we can move the exponent \( n \) in front of the logarithm, transforming it into a multiplicative coefficient. This step is particularly useful when faced with expressions like \( \log_{2}(xy)^{10} \), where an exponent is attached to a compound term. By applying the power rule, you simplify the problem to find:

• \[ \log_{2}((xy)^{10}) = 10 \log_{2}(xy) \]

Thus, you significantly simplify the expression by moving the exponent outside the logarithm, making the subsequent application of other logarithm rules easier. This rule is fundamental in converting complicated expressions into more manageable forms.

Product Rule of Logarithms

The product rule of logarithms comes into play when you have the logarithm of a product, and you want to split it into separate logarithmic terms. The formula that represents this is:

• \( \log_b(MN) = \log_b(M) + \log_b(N) \)

This rule is extremely helpful in breaking down complex terms involving multiplication, like the term \( \log_{2}(xy) \) into individual parts. By applying the product rule, we can convert:

• \[ 10 \log_{2}(xy) = 10 (\log_{2}(x) + \log_{2}(y)) \]

First, note that we need to identify what each part of the product \( MN \) corresponds to in the term \( xy \), so here \( M = x \) and \( N = y \). This allows you to expand and work with each component separately. The product rule effectively breaks down the problem, making it easier to solve by managing simpler logarithmic expressions. It is vital for expanding logarithmic expressions and gaining a clearer path to the solution.

Logarithmic Expansion

Logarithmic expansion involves using the rules of logarithms, such as the power and product rules, to break down complex logarithmic expressions into simpler components. It is a methodical approach to simplify expressions and often involves distributing coefficients and applying the rules strategically. In our original exercise, we began with:

• \( \log_{2}(xy)^{10} \)

After applying the power and product rules, the next part is distributing the coefficients to further expand and simplify:

• \[ 10 (\log_{2}(x) + \log_{2}(y)) = 10 \log_{2}(x) + 10 \log_{2}(y) \]

We see that the 10, initially brought down using the power rule, is multiplied across the sum resulting from the product rule. Each term in the parenthesis is treated individually, and the coefficient is distributed accordingly. This process results in a more straightforward expression that is often easier to interpret or further manipulate when solving algebraic equations. Logarithmic expansion is essential in mathematics as it translates complex structures into elementary forms. This, in turn, facilitates better insight and a clearer path towards solving equations or understanding the underlying patterns.