Problem 21 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \log _{6}(216 \sqrt{6}) $$

Step-by-Step Solution

Answer

The value of the logarithmic expression \(\log_{6}(216 \sqrt{6})\) is \(\frac{7}{2}\).

Step 1: Rewrite the expression inside the log

First, we need to rewrite \(216 \sqrt{6}\) in a more manageable form. Since \(6^3 = 216\) and \(\sqrt{6} = 6^{\frac{1}{2}}\), we can rewrite the expression as: \[ \log_{6}(216 \sqrt{6}) = \log_{6}(6^3 \cdot 6^{\frac{1}{2}}) \]

Step 2: Use exponent addition rule

Next, we'll use the exponent addition rule, which states that \(a^{m} \cdot a^{n} = a^{m + n}\). Applying this rule to our expression, we get: \[ \log_{6}(6^3 \cdot 6^{\frac{1}{2}}) = \log_{6}(6^{3 + \frac{1}{2}}) \]

Step 3: Simplify the exponent

Now, let's simplify the exponent: \[ 3 + \frac{1}{2} = \frac{6}{2} + \frac{1}{2} = \frac{7}{2} \] So, our expression becomes: \[ \log_{6}(6^{3 + \frac{1}{2}}) = \log_{6}(6^{\frac{7}{2}}) \]

Step 4: Use the property of logarithm

We can now use the property of logarithms that states \(\log_{a}(a^b) = b\). Applying this property to our expression, we obtain the final answer: \[ \log_{6}(6^{\frac{7}{2}}) = \frac{7}{2} \] Thus, the value of the given logarithmic expression is \(\frac{7}{2}\).

Key Concepts

Logarithm Simplification

Understanding logarithm simplification is crucial for solving complex problems with ease. A logarithm, expressed as \(\text{log}_{b}(x)\), essentially asks the question: 'To what power must we raise the base \(b\) to get \(x\)?' Simplification comes into play when the argument of the logarithm, which is \(x\) in our case, involves powers of the base.

For instance, if you have \(\text{log}_{6}(216 \times \text{6}^{1/2})\), simplifying it requires recognizing that \(216\) is a power of \(6\) and \(\text{6}^{1/2}\) is also a power of \(6\). By expressing the entire argument in terms of the base of the logarithm, the simplification process becomes straightforward. In this case, \(216\) is rewritten as \(6^{3}\), and the square root of \(6\) as \(6^{1/2}\). This step is essential to make use of other logarithm properties, which we will explore further.

Exponent Addition Rule

When dealing with powers of the same base within a logarithmic expression, you can simplify your work by applying the exponent addition rule. This rule states that when multiplying powers with the same base, you add the exponents: \(a^{m} \times a^{n} = a^{m+n}\).

This comes incredibly handy in logarithmic simplification because once you've expressed your terms with the same base, as we did with \(6^{3} \times 6^{1/2}\), you can simply add the exponents to consolidate the powers into a single term. This turns \(6^{3} \times 6^{1/2}\) into \(6^{7/2}\), streamlining the expression for the next steps of evaluation.

Logarithmic Expression Evaluation

To evaluate the logarithm of an expression where the argument is a power of the base, you use the fundamental log property \(\text{log}_{a}(a^b) = b\). This essentially strips away the logarithm, leaving you with the exponent alone since the base and the argument's base match perfectly.

In our example of \(\text{log}_{6}(6^{7/2})\), applying this logarithm property results in a direct answer: \(7/2\). This step concludes the evaluation, simplifying the original logarithmic term to the power itself. It's a process that shows the power of understanding logarithmic properties to break down and solve expressions that may at first seem complex.