Problem 53 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \log _{\sqrt{6}} 3 \cdot \log _{3} 36+\log _{\sqrt{3}} 8 \cdot \log _{4} 81 $$

Step-by-Step Solution

Answer

The simplified expression is: \( 13 + \log_{3} 2 \)

Step 1: Identify given expression

The given expression is: \[ \log _{\sqrt{6}} 3 \cdot \log _{3} 36+\log _{\sqrt{3}} 8 \cdot \log _{4} 81 \] Step 2: Simplify each logarithm

Step 2: Apply logarithm properties

We can apply the power rule, logarithm base change, and properties of exponents to each of the logarithms: \( \log_{\sqrt{6}} 3 \): Since \(3 = 6^{1/2}\sqrt{6}\), the expression becomes \(\log_{\sqrt{6}} (6^{1/2}\) \( \log_{3} 36 \): Since \(36 = 3^2\cdot 2^2\), the expression becomes \(\log_{3} (3^2\cdot 2^2) = 2\log_{3} 3 + 2\log_{3} 2\) \( \log_{\sqrt{3}} 8 \): Since \(8 = 2^3\), the expression becomes \( \log_{\sqrt{3}} 2^3\) \( \log_{4} 81 \): Since \(81 = 3^4\), the expression becomes \( \log_{4} 3^4\) Step 3: Use Change of Base Formula

Step 3: Apply Change of Base Formula

To make calculations easier, we can apply the Change of Base Formula and use the natural logarithm with base \(e\): (\( \log_{\sqrt{6}} 3 \cdot \log _{3} 36 \)): \[ \frac{\ln (6^{1/2})}{\ln (\sqrt{6})} \cdot (2\log_{3} 3 + 2\log_{3} 2) = \frac{1}{2}\cdot (2 + 2\log_{3} 2) \] (\( \log_{\sqrt{3}} 8 \cdot \log_{4} 81 \)): \[ \frac{\ln (2^3)}{\ln (\sqrt{3})} \cdot \frac{\ln (3^4)}{\ln 4} = \frac{3}{\frac{1}{2}}\cdot \frac{4}{2} = 6\cdot 2 \] Step 4: Combine results

Step 4: Combine simplification results

Now, we can combine our results from steps 2 and 3: \[ \frac{1}{2}(2 + 2\log_{3} 2) + 6\cdot 2 \] Step 5: Simplify the expression

Step 5: Simplify the final expression

Finally, we can simplify the expression: \[ \frac{1}{2}(2 + 2\log_{3} 2) + 12 = 1 + \log_{3} 2 + 12 = 13 + \log_{3} 2 \] So, the simplified expression is: \[ 13 + \log_{3} 2 \]

Key Concepts

Logarithm Properties

Understanding the properties of logarithms is pivotal to simplifying complex logarithmic expressions. In essence, logarithms are exponents, representing the power to which a base number must be raised to produce a certain value.

One crucial property is the power rule, which states that \( \log_b(m^n) = n \log_b(m) \), essentially allowing us to move the exponent of the argument out in front of the log. This is especially useful when the base of the log and the argument (the number inside the log function) have a clear exponential relationship, like \( \log_{\sqrt{6}} 3 \), where \(3\) is related to \(6\) via an exponent of \(1/2\).

Another property demonstrated in the solution is the product rule. This property indicates that the logarithm of a product, such as \( \log_b(m \cdot n) \), can be expressed as the sum of two logs: \( \log_b(m) + \log_b(n) \). This property was applied to \( \log_{3} 36 \) by breaking down \(36\) into \(3^2 \cdot 2^2\), simplifying it to \(2\log_{3} 3 + 2\log_{3} 2\).

These properties highlight the interconnectedness of algebra and logarithms, allowing us to break down and reconstruct logarithmic expressions to more manageable forms.

Change of Base Formula

When faced with logs with bases that are not common or practical for computation, the change of base formula becomes an indispensable tool. This formula allows us to convert a log of any base into a log with a base of our choosing, typically to the natural logarithm \( \log_e \), also known as \( \ln \), or the common logarithm \( \log_{10} \).

The formula is stated as follows: \( \log_b(a) = \frac{\log_c(a)}{\log_c(b)} \), where \( b \) is the original base, \( a \) is the argument, and \( c \) is the new base. In the exercise solution, this formula was used to convert all logs into natural logarithms, making use of the mathematical relationships simpler. For example, \( \log_{\sqrt{6}} 3 \) was expressed as \( \frac{\ln (6^{1/2})}{\ln (\sqrt{6})} \).

The decision to employ the natural logarithm in the change of base formula is often guided by its prevalence in calculus and its inherent simplicity, thanks in part to the number \( e \), which serves as its base and appears frequently in continuous growth models and other aspects of mathematics.

Logarithmic Expressions

Logarithmic expressions encapsulate any expression that involves logarithms. These can range from a single log function to more complex expressions involving multiple logs with different bases, products, or quotients, as seen in the exercise.

To manipulate these expressions effectively, one must combine a deep understanding of logarithm properties with algebraic manipulation skills. This meld allows the simplification of even the most daunting expressions.

Understanding and Simplifying

The initial complex expression, \( \log _{\sqrt{6}} 3 \cdot \log _{3} 36+\log _{\sqrt{3}} 8 \cdot \log _{4} 81 \), could be intimidating at first. However, by breaking it down using the properties of logarithms and the change of base formula, it became much more approachable.

Once broken down into simpler components and converted to a common base where necessary, the final steps involved combining and simplifying the results using basic arithmetic to yield \( 13 + \log_{3} 2 \). A clear, methodical approach to logarithmic expressions, as illustrated, ensures that students can not only arrive at the correct answer but also understand the 'why' and 'how' behind each step.