Problem 24 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

If \(\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}\), then prove that \(a^{a} b^{b} \cdot c^{c}=1\)

Step-by-Step Solution

Answer

By applying the property of logarithms and combing the equations, we have proved that \(a^{a} b^{b} c^{c}=1\)

Step 1: Apply the Property of Logarithms

Noting that the three ratios are equal, we can set a common variable \( k \) to represent them. So, we have \(\log a = k(b - c)\), \(\log b = k(c - a)\), and \(\log c = k(a - b)\)

Step 2: Combine the equations

Add all three equations to simplify our system: \(\log a + \log b + \log c = k(b - c + c - a + a - b) = 0\). Using properties of logarithms, the left side of this equation can be simplified to \(\log (abc) = 0\)

Step 3: Solve the result

We know that \(\log (abc) = 0\) implies \(abc = 1\) as \(\log 1 = 0\) in any base. Thus, we have \( a^{a} b^{b} c^{c} = 1^{(a+b+c)} = 1 \) because when we raise 1 to any power, we always get 1. So we've arrived at the result.

Key Concepts

Properties of Logarithms

To understand how to solve logarithmic equations like the one in the original exercise, it is important to grasp the properties of logarithms. Logarithms are the inverses of exponentiation and have specific rules that make them manipulatable in algebraic expressions.

One essential property is the 'product rule', which states that the logarithm of a product is equal to the sum of the logarithms of the individual factors. In mathematical terms, this is expressed as \( \log(ab) = \log a + \log b \). This property is used to combine the logarithms in the exercise and simplify the equation.

Another important property is the 'zero rule', which states \( \log_b 1 = 0 \) for any base \( b \). The rule is pivotal in the second step where \( \log (abc) = 0 \) simplifies to \( abc = 1 \), following the definition that any base raised to the power of zero equals one.

Exponential Equations

An exponential equation is one in which a variable appears in the exponent. Solving these types of equations often involves isolating the variable and using properties of exponents and logarithms to do so.

In the exercise, we encounter a subtle form of an exponential equation when we say \( a^{a} b^{b} c^{c} = 1^{(a+b+c)} \). This can be considered exponential because the variables act as both the base and the exponent. Here, the student needs to recognize that raising anything to the power of zero gives one, which aligns with solving a basic exponential equation \( b^0 = 1 \) where \( b \) can be any non-zero number.

The power of algebraic properties shines through in the way we deduce the relationship from logarithmic to exponential form, applying the insight that \( \log_b x = y \) is equivalent to \( b^y = x \) to reach the solution.

Trigonometry Problem Solving

While the given problem does not directly involve trigonometry, the approach to solving complex equations using transformation and substitution is a skill also valuable in tackling trigonometry problems. Trigonometric problem solving often involves using identities to transform the original equations into a more workable form, similar to how we manipulate logarithmic equations.

In trigonometry, these techniques become essential when dealing with equations involving sine, cosine, and tangent functions. For example, one might use the Pythagorean identity to substitute \( \sin^2(x) \) with \( 1 - \cos^2(x) \), if this simplifies the problem at hand.

Thus, the logical reasoning and algebraic skills honed through solving logarithmic and exponential equations lay a solid foundation for success with trigonometry problem solving. Mastery of these skills helps students to approach mathematical problems with confidence and inventiveness.