Problem 69 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

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

Step-by-Step Solution

Answer

To prove that \(\log_b a \cdot \log_c b \cdot \log_a c = 1\), we apply the change of base formula and rewrite the logarithms in terms of natural logarithms (base e): \(\frac{\ln a}{\ln b} \cdot \frac{\ln b}{\ln c} \cdot \frac{\ln c}{\ln a} = 1\). Simplifying the equation by canceling out terms gives us \(\frac{1}{1} = 1\), confirming that the initial equation holds true.

Step 1: Write down the given equation

The given equation is: \(\log_b a \cdot \log_c b \cdot \log_a c = 1\)

Step 2: Apply the Change of Base Formula

We use the change of base formula to rewrite the logarithms in terms of the natural logarithm (base e). The formula states: \(\log_x y = \frac{\ln y}{\ln x}\) First, we rewrite the logarithms using the change of base formula: \(\frac{\ln a}{\ln b} \cdot \frac{\ln b}{\ln c} \cdot \frac{\ln c}{\ln a} = 1\)

Step 3: Simplify the Equation

Now, we will simplify the equation by canceling out terms. Observe that there are pairs of numerators and denominators that cancel, like: * \(\ln a\) in the numerator of the first fraction and the denominator of the last fraction * \(\ln b\) in the numerator of the second fraction and the denominator of the first fraction * \(\ln c\) in the numerator of the last fraction and the denominator of the second fraction So, the resulting equation simplifies to: \(\frac{1}{1} = 1\)

Step 4: Conclusion

Since the simplified equation results in 1, we've successfully proven that: \(\log_b a \cdot \log_c b \cdot \log_a c = 1\)

Key Concepts

Change of Base Formula

The change of base formula is a nifty little tool that helps you convert logarithms from one base to another. This is particularly useful when you're working with different bases and prefer to keep things simple with a common base, like the natural logarithm (base \( e \)).

The formula to change the base of a logarithm is:

• \( \log_x y = \frac{\ln y}{\ln x} \)

This means you can express the logarithm of \( y \) to any base \( x \) as the ratio of the natural logarithm of \( y \) to the natural logarithm of \( x \).

This technique comes in handy to simplify complex expressions, like in our original exercise. By converting each logarithm into a ratio of natural logarithms, it makes the cancellation of terms straightforward and proves the equation easily. The change of base formula is fundamental for handling logarithms in practical scenarios.

Simplifying Logarithmic Expressions

Simplifying logarithmic expressions involves breaking down complex logarithmic terms into simpler forms. This process often uses properties of logarithms and the change of base formula to make calculations manageable.

Consider the expression in our example: \( \log_b a \cdot \log_c b \cdot \log_a c \). Initially, this might seem like a tough nut to crack, but with the change of base formula, it simplifies beautifully!

By rewriting each term in the expression as a fraction of natural logarithms, as in:

• \( \frac{\ln a}{\ln b} \cdot \frac{\ln b}{\ln c} \cdot \frac{\ln c}{\ln a} \)

You can easily cancel out similar terms across fractions:

• \( \ln a \) cancels with \( \ln a \)
• \( \ln b \) cancels with \( \ln b \)
• \( \ln c \) cancels with \( \ln c \)

This leaves you with \( \frac{1}{1} = 1 \), demonstrating how simplifying can yield dramatic results. This skill can make dealing with logarithmic expressions less tedious.

Logarithm Properties

Logarithms have several properties that are invaluable when manipulating expressions. These properties make simplifying expressions and solving equations much easier.

Some key properties of logarithms include:

• The Product Rule: \( \log_b (xy) = \log_b x + \log_b y \)
• The Quotient Rule: \( \log_b \left(\frac{x}{y}\right) = \log_b x - \log_b y \)
• The Power Rule: \( \log_b (x^y) = y \cdot \log_b x \)
• The Change of Base Formula: \( \log_x y = \frac{\ln y}{\ln x} \)

These rules allow you to transform and simplify expressions by breaking them down into more manageable parts.

In proofs like the one in our example, properties of logarithms enable you to rework the expressions strategically. For instance, changing bases or identifying patterns can be crucial steps as you look to verify identities or solve equations. Familiarity with these properties paves the way to solving a wide range of problems with ease.