Problem 38 - Algebra and Trigonometry

Question

Use the Laws of Logarithms to expand the expression. $$ \log \left(\frac{10^{x}}{x\left(x^{2}+1\right)\left(x^{4}+2\right)}\right) $$

Step-by-Step Solution

Answer

The expanded expression is: \(x - \log(x) - \log(x^2 + 1) - \log(x^4 + 2)\).

Step 1: Apply the Quotient Rule

The first step is to apply the quotient rule of logarithms, which states that \( \log\left(\frac{a}{b}\right) = \log(a) - \log(b) \). In this expression, let \( a = 10^{x} \) and \( b = x(x^2 + 1)(x^4 + 2) \). Thus, we have:\[ \log\left(\frac{10^{x}}{x(x^2+1)(x^4+2)}\right) = \log(10^{x}) - \log\left(x(x^2 + 1)(x^4 + 2)\right). \]

Step 2: Simplify the Logarithm of a Power

For the term \( \log(10^{x}) \), use the power rule of logarithms, which states that \( \log(a^b) = b \cdot \log(a) \). Here, \( a = 10 \) and \( b = x \). Therefore, we have:\[ \log(10^{x}) = x \cdot \log(10). \] Since \( \log(10) = 1 \), this simplifies to \( x. \)

Step 3: Apply the Product Rule

Next, apply the product rule of logarithms to the expression \( \log\left(x(x^2 + 1)(x^4 + 2)\right) \). According to the product rule, \( \log(abc) = \log(a) + \log(b) + \log(c) \). Let \( a = x \), \( b = (x^2 + 1) \), and \( c = (x^4 + 2) \). Thus, we have:\[ \log\left(x(x^2 + 1)(x^4 + 2)\right) = \log(x) + \log(x^2 + 1) + \log(x^4 + 2). \]

Step 4: Combine All Parts

Combine all parts to get the expanded expression. Start with \( x \), then subtract the expanded form of the denominator from Step 3:\[ \log\left(\frac{10^{x}}{x(x^2+1)(x^4+2)}\right) = x - \log(x) - \log(x^2 + 1) - \log(x^4 + 2). \]

Key Concepts

Quotient Rule

The quotient rule is an essential concept in logarithms that often helps simplify complex expressions. It states that for any two positive numbers, say \( a \) and \( b \), the logarithm of their division can be written as: \[ \log\left(\frac{a}{b}\right) = \log(a) - \log(b). \]
This can be incredibly useful as it allows a division within a logarithmic function to be broken down into a subtraction operation between two separate logarithmic expressions. Let's illustrate this with an example from our exercise:

• We start with \( a = 10^x \) and \( b = x(x^2+1)(x^4+2) \).
• Using the quotient rule, the expression becomes: \[ \log(10^x) - \log(x(x^2 + 1)(x^4 + 2)). \]

The quotient rule makes tackling otherwise daunting logarithmic expressions much easier by converting a division into a simpler subtraction problem.

Power Rule

The power rule in logarithms is another cornerstone concept. When you have a logarithmic expression with an exponent, like \( a^b \), you can use the power rule to simplify it to: \[ \log(a^b) = b \cdot \log(a). \]
The benefit of this rule lies in how it transforms a more complex exponential form into a straightforward multiplication. For instance, in our current exercise:

• \( a = 10 \) and \( b = x \).
• Thanks to the power rule, \( \log(10^x) = x \cdot \log(10). \)
• Since \( \log(10) = 1 \), the expression simplifies further to just \( x \).

This rule is particularly beneficial when dealing with exponential components as it essentially "brings down" the exponent, simplifying multiplication inside the log.

Product Rule

The product rule for logarithms helps decompose a logarithm of a product into a sum of separate logs. This rule states that for any three positive numbers \( a, b, \) and \( c \), you can write: \[ \log(abc) = \log(a) + \log(b) + \log(c). \]
This is particularly helpful when you encounter a logarithm applied to a multiplication, as it allows you to handle each part separately. In the example provided:

• In \( \log(x(x^2+1)(x^4+2)) \), consider \( a = x \), \( b = x^2+1 \), \( c = x^4+2 \).
• The product rule lets us break this down as: \[ \log(x) + \log(x^2 + 1) + \log(x^4 + 2). \]

Each term is now an individual logarithmic expression to work with, making complex expressions far more manageable. The product rule is handy when you need to separate intertwined multiplicative components within a logarithmic formula.