Problem 35 - Algebra and Trigonometry

Question

\(19-44\) Use the Laws of Logarithms to expand the expression. $$ \log _{2}\left(\frac{x\left(x^{2}+1\right)}{\sqrt{x^{2}-1}}\right) $$

Step-by-Step Solution

Answer

\( \log_2(x) + \log_2(x^2+1) - \frac{1}{2} \log_2(x^2-1) \)

Step 1: Apply the Quotient Rule

The logarithmic expression is given as \( \log_2 \left( \frac{x(x^2+1)}{\sqrt{x^2-1}} \right) \). According to the quotient rule of logarithms, \( \log_b \left( \frac{M}{N} \right) = \log_b M - \log_b N \). Here, \( M = x(x^2+1) \) and \( N = \sqrt{x^2-1} \). So, we write:\[ \log_2 \left( \frac{x(x^2+1)}{\sqrt{x^2-1}} \right) = \log_2 (x(x^2+1)) - \log_2 (\sqrt{x^2-1}) \]

Step 2: Expand the Numerator with the Product Rule

The expression in the numerator \( x(x^2+1) \) can be expanded using the product rule of logarithms, which states \( \log_b (MN) = \log_b M + \log_b N \). Let's apply it here:\[ \log_2(x(x^2+1)) = \log_2(x) + \log_2(x^2+1) \]

Step 3: Simplify the Denominator with the Power Rule

The square root in the denominator \( \sqrt{x^2-1} \) can be expressed as \( (x^2-1)^{1/2} \). According to the power rule \( \log_b (M^k) = k \log_b M \), we write:\[ \log_2(\sqrt{x^2-1}) = \frac{1}{2} \log_2(x^2-1) \]

Step 4: Compile the Expanded Expression

Putting all the pieces together from Step 1, 2, and 3, the expanded logarithmic expression is:\[\log_2 \left( \frac{x(x^2+1)}{\sqrt{x^2-1}} \right) = \log_2(x) + \log_2(x^2+1) - \frac{1}{2} \log_2(x^2-1) \]

Key Concepts

Quotient Rule of Logarithms

When dealing with logarithmic expressions, the quotient rule is a powerful tool. This rule helps simplify a logarithm that involves division. Let's break it down: the logarithm of a quotient is the difference between the logarithms of the numerator and the denominator. Mathematically, it is written as:

• \( \log_b \left( \frac{M}{N} \right) = \log_b M - \log_b N \)

In the given expression, \( M = x(x^2+1) \) and \( N = \sqrt{x^2-1} \). So, applying the quotient rule allows us to separate the original expression into two simpler logarithmic forms. This makes solving or expanding logarithmic expressions much more manageable.
Think of it as "unpacking" the logarithm to deal with each part individually. The new expression, as a result, becomes:

• \( \log_2 \left( \frac{x(x^2+1)}{\sqrt{x^2-1}} \right) = \log_2 (x(x^2+1)) - \log_2 (\sqrt{x^2-1}) \)

Once unpacked, you can tackle each element (numerator and denominator) with other rules, such as the product or power rule, to further expand or simplify.

Product Rule of Logarithms

The product rule for logarithms is all about breaking down products within a logarithm into a sum. This is especially helpful when the expression is complicated by multiplication inside the log.The product rule is expressed as:

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

Take the expression from the previous section, \( x(x^2+1) \), found in the numerator.
Here, by applying the product rule, we quickly identify it as a product of \( x \) and \( x^2+1 \). Thus, the expression expands to:

• \( \log_2 (x(x^2+1)) = \log_2(x) + \log_2(x^2+1) \)

Using this rule gives us a breakdown, itemizing each part of the original product, so you can deal with each term on its own. That's how the product rule simplifies complex expressions, making each piece more approachable.

Power Rule of Logarithms

Handling powers within logarithms can be simplified using the power rule. This rule allows you to move exponents out in front of the logarithm, making expressions easier to work with.The power rule states:

• \( \log_b (M^k) = k \log_b M \)

In our expression, the denominator has a square root: \( \sqrt{x^2-1} \). Remember that the square root can be rewritten as an exponent, specifically as \((x^2-1)^{1/2}\).
Applying the power rule here lets us extract the fraction exponent, making the expression:

• \( \log_2(\sqrt{x^2-1}) = \frac{1}{2} \log_2(x^2-1) \)

This step is invaluable because it reduces complex calculations with roots or higher powers inside a logarithm by transforming them into multiplication outside.
Moving the exponent to the front means fewer complications and more straightforward calculations with logarithms, ensuring your work isn't bogged down by unnecessary complexity.