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Chapter 1: Problem 70

$$ \text { Prove that } \log _{b} a \log _{c} b \log _{d} c \log _{a} d=1 $$

Short Answer

Expert verified
Using the properties of logarithms, particularly the Change of Base Formula, we can rewrite the given expression and simplify it by cancelling out terms: \[ \frac{\log_c a}{\log_c b} \cdot \frac{\log_d b}{\log_d c} \cdot \frac{\log_a c}{\log_a d} \cdot \frac{\log_b d}{\log_b a} = 1 \] After cancelling out all terms and simplifying, we achieve the desired result: \[ \log _{b} a \log _{c} b \log _{d} c \log _{a} d = 1 \]

Step by step solution

01

Recognize the logarithm properties

Before diving into solving the expression, it's important to familiarize ourselves with the properties of logarithms. The properties that will be useful for solving this exercise are: 1. Change of Base Formula: \[ \log_b a = \frac{\log_c a}{\log_c b} \] 2. Logarithmic identities: \[ \log_b 1 = 0 \quad \text{and} \quad \log_b b = 1 \] Now, we will use these properties to solve the given expression.
02

Use the change of base formula to represent the given expression

Let's use the change of base formula for each of the logarithms in the given expression: \[ \log_b a = \frac{\log_c a}{\log_c b} \] \[ \log_c b = \frac{\log_d b}{\log_d c} \] \[ \log_d c = \frac{\log_a c}{\log_a d} \] \[ \log_a d = \frac{\log_b d}{\log_b a} \] Now, let's substitute these expressions back into the given equation: \[ \frac{\log_c a}{\log_c b} \cdot \frac{\log_d b}{\log_d c} \cdot \frac{\log_a c}{\log_a d} \cdot \frac{\log_b d}{\log_b a} = 1 \]
03

Simplify the expression

Now, we'll simplify the expression by cancelling out the terms: \[ \frac{\log_c a}{\cancel{\log_c b}} \cdot \frac{\cancel{\log_d b}}{\cancel{\log_d c}} \cdot \frac{\cancel{\log_a c}}{\cancel{\log_a d}} \cdot \frac{\cancel{\log_b d}}{\cancel{\log_b a}} = 1 \] After cancelling out all terms, we are left with the following: \[ \frac{\log_c a}{1} \cdot \frac{1}{1} \cdot \frac{1}{1} \cdot \frac{1}{\log_b a} = 1 \] Now, multiply the expression: \[ \frac{\log_c a}{\log_b a} = 1 \] Finally, notice that both numerator and denominator are equal by the change of base formula: \[ \log_b a = \log_b a \] So, we can conclude that the given expression equals 1: \[ \log _{b} a \log _{c} b \log _{d} c \log _{a} d = 1 \] The proof is complete.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Change of Base Formula
The change of base formula is a powerful tool in logarithms, which allows the conversion of a logarithm from one base to another. This formula is crucial for simplifying and solving logarithmic expressions.

The change of base formula can be expressed as:
  • \[\log_b a = \frac{\log_c a}{\log_c b}\]
Here, \(b\) and \(c\) are the bases, and the formula shows how to convert \(\log_b a\) into a logarithmic expression with base \(c\). This can be very handy when you are working on calculations where the base isn鈥檛 very convenient. Most modern calculators only allow logarithms in base 10 (common logarithm) or base \(e\) (natural logarithm). Using the change of base formula, you can convert any logarithm to these available bases, which makes calculations much easier to manage.

By utilizing this formula, the original complex expression can be broken down into simpler fractions of logarithms, making it easier to simplify further and solve.
Properties of Logarithms
The properties of logarithms form the foundation for manipulating and simplifying logarithmic expressions. Understanding these properties can provide deeper insights into how logarithms work:
  • Product Property: \(\log_b (MN) = \log_b M + \log_b N\)
  • Quotient Property: \(\log_b \left( \frac{M}{N} \right) = \log_b M - \log_b N\)
  • Power Property: \(\log_b (M^n) = n \log_b M\)
  • Identity Property: \(\log_b b = 1\)
  • Zero Property: \(\log_b 1 = 0\)
These properties are fundamental when it comes to transforming or simplifying logarithmic equations. In our particular problem, the identity property (\(\log_b b = 1\)) and the zero property (\(\log_b 1 = 0\)) helped simplify the fractions and the expression itself.

Applying these properties can significantly reduce the complexity of problems, allowing for easier problem-solving processes. By recognizing how each property can be applied, you can become much more proficient in working with logarithms.
Logarithmic Equations
Logarithmic equations involve equations that include logarithms, and solving them requires a keen understanding of the properties and potentially the change of base formula. Solving these equations often means isolating the logarithmic expression on one side of the equation and transforming the equation into an exponential form. With logarithmic equations, here are some strategies you might find useful:
  • Use properties of logarithms to combine or break apart logs to simplify the equation.
  • If possible, convert the logarithmic equation to the same base to easily solve for the variable.
  • Remember that you can also exponentiate both sides of an equation to remove the logs completely, turning the equation into an exponential one.
In the exercise provided, we used a combination of logarithm properties and the change of base formula to ultimately simplify and solve the equation, proving the product of four logarithms equals one.
Understanding these strategies can greatly enhance your problem-solving skills when dealing with more complicated logarithmic equations, helping you to efficiently navigate through the complexity of these problems.

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Most popular questions from this chapter

$$ \log _{3} 2 \cdot \log _{4} 3 \cdot \log _{5} 4 \cdot \log _{6} 5 \cdot \log _{7} 6 \cdot \log _{8} 7 $$

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$$ \text { Find } \phi(\psi(x)) \text { and } \psi(\phi(x)) \text { if } \phi(x)=x^{2}+1 \text { and } \psi(x)=3^{x} \text { . } $$
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