Problem 70 - Precalculus with Calculus Previews

Question

In Problems \(69-72\), use the laws of logarithms in Theorem \(6.2 .1\) so that \(\ln y\) contains no products, quotients, or powers. $$ y=\sqrt{\frac{(2 x+1)(3 x+2)}{4 x+3}} $$

Step-by-Step Solution

Answer

\( \ln y = \frac{1}{2} \ln(2x+1) + \frac{1}{2} \ln(3x+2) - \frac{1}{2} \ln(4x+3) \)

Step 1: Rewrite the Square Root as a Power

First, express the square root as a fractional power. We have \( y = \left( \frac{(2x+1)(3x+2)}{4x+3} \right)^{1/2} \) .

Step 2: Apply the Power Rule for Logarithms

Use the power rule \( \ln(a^b) = b \cdot \ln(a) \). Thus, \( \ln y = \frac{1}{2} \ln \left( \frac{(2x+1)(3x+2)}{4x+3} \right) \).

Step 3: Expand the Logarithm of a Quotient

Use the quotient rule for logarithms \( \ln(\frac{a}{b}) = \ln a - \ln b \). Apply this to get \( \ln y = \frac{1}{2} (\ln((2x+1)(3x+2)) - \ln(4x+3)) \).

Step 4: Expand the Logarithm of a Product

Use the product rule \( \ln(a \cdot b) = \ln a + \ln b \). Therefore, \( \ln((2x+1)(3x+2)) = \ln(2x+1) + \ln(3x+2) \). Substituting back, we have \( \ln y = \frac{1}{2} (\ln(2x+1) + \ln(3x+2) - \ln(4x+3)) \).

Step 5: Simplify the Expression

Distribute the \( \frac{1}{2} \) into each term: \( \ln y = \frac{1}{2} \ln(2x+1) + \frac{1}{2} \ln(3x+2) - \frac{1}{2} \ln(4x+3) \). This expression for \( \ln y \) contains no products, quotients, or powers.

Key Concepts

Power Rule for Logarithms

The power rule for logarithms is a handy tool when dealing with exponents in logarithmic expressions. It states that the logarithm of a power, such as \( a^b \), can be simplified by bringing the exponent out in front as a multiplier. This means \( \ln(a^b) = b \cdot \ln(a) \).
To illustrate, consider the expression \( y = (\text{something})^{1/2} \), like in our exercise. When you take the natural logarithm of both sides, you use the power rule to simplify \( \ln(y) = \frac{1}{2} \cdot \ln(\text{something}) \).
Here's why it's beneficial:

• The rule allows you to handle complex powers easily.
• It turns a potentially complicated expression into a simpler form, making subsequent operations more straightforward.

In our problem, we applied this rule to move the exponent \( \frac{1}{2} \) in front, simplifying our logarithm considerably.

Quotient Rule for Logarithms

The quotient rule for logarithms provides an efficient way to separate a logarithm of a quotient into individual components. It is expressed as \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \).
This rule is vital when you want to break down logs of fractions into more manageable terms. In our exercise, the expression within the logarithm was a fraction, hinting at the perfect opportunity to use the quotient rule.
Let's see what happens when we apply it:

• We can decompose \( \ln\left(\frac{(2x+1)(3x+2)}{4x+3}\right) \) into \( \ln((2x+1)(3x+2)) - \ln(4x+3) \).
• This simplifies the problem by reducing the complexity of handling a single logarithmic expression and separating it into two simpler logarithms.

The quotient rule is especially powerful when dealing with longer algebraic fractions in logarithms, making each component easier to handle individually.

Product Rule for Logarithms

The product rule for logarithms is an essential property when dealing with multiplication within logarithmic expressions. It states that \( \ln(a \cdot b) = \ln(a) + \ln(b) \).
This rule is instrumental in our exercise, where the expression inside the logarithm consisted of a product. By applying the product rule, we convert this log into a sum of simpler logarithms.
Here's how it works:

• For an expression like \( \ln((2x+1)(3x+2)) \), using the product rule, we can separate it into \( \ln(2x+1) + \ln(3x+2) \).
• This breakdown means we handle smaller, less complicated terms, making calculation and simplification much easier.

The product rule not only splits complex products into more manageable parts, but it also complements other rules. In this problem, using both the product and quotient rules allows us to fully decompose and simplify the original logarithmic expression efficiently.