Problem 37 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

If \(a=\log _{12} 18\) and \(b=\log _{24} 54\), then prove that \(a b+5(a-b)=1\)

Step-by-Step Solution

Answer

The provided equation \(a b+5(a-b)=1\) is not valid for \(a=\log _{12} 18\) and \(b=\log _{24} 54\).

Step 1: Evaluate log values

Substitute \(a=\log _{12} 18\) and \(b=\log _{24} 54\) with the logarithm property: log (m/n) = log (m) - log (n). So, \(a = \log _{12} 18 = \log_{12}(2*9) = \log_{12}(2) + \log_{12}(9)\) and \(b = \log _{24} 54 = \log_{24}(2*27) = \log_{24}(2) + \log_{24}(27)\).

Step 2: Convert logarithm base

Convert to the same base for easier calculation. For a, convert \(\log_{12}2\) to \(\log_{4}2\), and \(\log_{12}9\) to \(\log_{4}9\). Similarly for b, convert \(\log_{24}2\) to \(\log_{8}2\), and \(\log_{24}27\) to \(\log_{8}27\).

Step 3: Further simplify log values

Solve these calculated log values. For a, \(\log_{4}2 = 0.5, \log_{4}9 = 1.5\), therefore \(a = 0.5 + 1.5 = 2\). For b, \(\log_{8}2 = 0.333..., \log_{8}27 = 2\), therefore \(b = 0.333... + 2 = 2.333...\).

Step 4: Substitute in equation

Substitute the values of a and b into the equation \(a b+5(a-b)=1\). This comes out as \(2 * 2.333... +5(2- 2.333...) = 1\).

Step 5: Simplify the equation

Solve the equation. The left hand side simplifies to \(4.666... -1.666... = 3\). Here, we see that the sides are not equal, hence, the provided equation was incorrect.

Key Concepts

Logarithm Properties

Understanding the properties of logarithms is essential for solving logarithmic equations. Logarithms are the inverse operations of exponentiation. They help in tackling problems involving exponential calculations by simplifying the operations. Here are some key properties:

• Product Rule: The logarithm of a product is the sum of the logarithms of the factors, i.e., \( \log_b (xy) = \log_b x + \log_b y \).
• Quotient Rule: The logarithm of a quotient is the difference between the logarithm of the numerator and the logarithm of the denominator, i.e., \( \log_b \left( \frac{x}{y} \right) = \log_b x - \log_b y \).
• Power Rule: The logarithm of a power is the exponent times the logarithm of the base, i.e., \( \log_b (x^n) = n \cdot \log_b x \).
• Change of Base Formula: This allows converting a logarithm to a different base, which is particularly useful when calculating non-standard bases: \( \log_b a = \frac{\log_k a}{\log_k b} \), where \( k \) is any positive number.

These properties allow us to manipulate and solve complex logarithmic equations efficiently. In the original exercise, the use of these properties was crucial to break down the given expressions \( \log_{12}18 \) and \( \log_{24}54 \) into simpler forms.

Base Conversion in Logarithms

Switching the base of logarithms can simplify the computation significantly. This process is often used to make calculations easier by converting logs to a base for which it is easier to compute or already known.To convert between bases, we use the Change of Base Formula:

• This formula is given by \( \log_b a = \frac{\log_k a}{\log_k b} \), allowing us to express a logarithm in any desired base \( k \).
• In practice, this often means converting to base 10 or base \( e \) (natural logarithm), as these can typically be easily calculated using calculators.

For example, in the original solution, the logarithms \( \log_{12} 18 \) and \( \log_{24} 54 \) were expressed in base 4 and base 8, respectively, to simplify and arrive at numerical approximations swiftly.In the step-by-step breakdown, converting to these intermediary bases allows leveraging known log values like \( \log_{4} 2 \) or \( \log_{8} 2 \), streamlining the calculation process.

Trigonometry for Competitive Exams

While the current exercise focuses on logarithmic equations, understanding both logarithms and trigonometry is vital in competitive exams like SAT, ACT, and various forms of mathematics olympiads. Trigonometric functions often interplay with logarithms in problems that involve growth, oscillation, or approximation, making expertise in both areas useful.Here’s how mastering trigonometry helps:

• Angle Functions: Know the basics of sine, cosine, and tangent functions. These form the foundation for solving equations involving angles and distances.
• Identities: Trigonometric identities, like \( \sin^2 \theta + \cos^2 \theta = 1 \), are used to simplify complex expressions.
• Application in Logs: Occasionally, logs and trig functions are combined in problems, especially in calculus-based tasks.

Often, students might find questions structured in a way that requires effective use of both logarithmic knowledge and trigonometric insight to solve.Preparing for exams involve familiarizing oneself not only with formulas but also learning to apply these interrelated math concepts in problem-solving scenarios.