Problem 64 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { Given that } \log _{36} 8=a, \text { find } \log _{36} 9 \text { in terms of } a \text { . } $$

Step-by-Step Solution

Answer

In terms of \(a\), \(\log_{36}9 = 0\).

Step 1: Rewrite the given expression in exponential form

We can rewrite the given equation \(\log_{36}8 = a\) in the exponential form: \[36^a = 8\]

Step 2: Use the change of base formula for logarithms

To express \(\log_{36}9\) in terms of \(a\), we can use the change of base formula: \(\log_{36}9 = \frac{\log_b 9}{\log_b 36}\), where \(b\) is any positive number, and \(b \neq 1\).

Step 3: Choose a convenient base for the logarithms

We can choose the base \(b = 2\), since both numbers \(9\) and \(36\) can easily be expressed as powers of \(2\): \(\log_{36}9 = \frac{\log_2 9}{\log_2 36}\)

Step 4: Simplify the logarithms

Now, we can express both \(9\) and \(36\) as powers of \(2\): \(\log_{36}9 = \frac{\log_2 (3^2)}{\log_2 (6^2)} = \frac{2 \log_2 3}{2 \log_2 6}\) Since \(2\) is a common factor in numerator and denominator, we can cancel it out: \(\log_{36}9 = \frac{\log_2 3}{\log_2 6}\)

Step 5: Relate the expression to the given variable \(a\)

Recall from Step 1, the given equation: \(36^a = 8\) Rewrite \(8\) and \(36\) as powers of \(2\): \(2^{4a} = 2^3\) Since the base is the same, we can equate their exponents: \(4a = 3\) Now, we can solve for \(a\): \(a = \frac{3}{4}\)

Step 6: Substitute the value of \(a\) into the expression

Now, we will express \(\log_2 3\) in terms of \(a\): \(\log_2 3 = 3 - \log_2 6 = 3 - \frac{3}{4} \cdot 4 = 3 - 3 = 0\) Then, substitute this into the expression we derived for \(\log_{36}9\): \(\log_{36}9 = \frac{\log_2 3}{\log_2 6} = \frac{0}{\log_2 6} = 0\) In conclusion, we found that in terms of \(a\), \(\log_{36}9 = 0\).

Key Concepts

Change of Base Formula

The change of base formula is a powerful tool for working with logarithms. It allows you to express a logarithm from one base in terms of another base. This formula is particularly helpful when you need to simplify or calculate logarithms that aren't easy to evaluate with standard bases like 10 or 2.
To apply the change of base formula, consider a logarithm \(\log_b a\). The formula is given by:

• \( \log_b a = \frac{\log_k a}{\log_k b}\)

Where \(k\) is a new base, and can be any positive number different from 1. Commonly used bases are 10 (common logarithms) or e (natural logarithms). This formula essentially changes the computation of a logarithm to one that might be easier, by potentially using calculators capable of computing logarithms with these bases.
The change of base formula was used in the solution to convert \(\log_{36} 9\) into a form that could be calculated using logarithms of base 2. This simplification made it easier to find the final result.

Exponential Form

Understanding exponential form is a cornerstone of grasping the concept of logarithms. At its core, logarithms are the inverse operations of exponentiation. The relationship between logarithms and exponentials is fundamental and easy to grasp once you see how they are inversely connected.
When you have a logarithmic expression such as \(\log_{b} a = x\), it can be rewritten in exponential form as:

• \(b^x = a\)

This transformation shows how exponentiation and logarithms are related. An exponential expression is basically a statement that the base raised to some power results in a particular number. So whenever you're working with logarithms, you can always translate them into their equivalent exponential form to see the relationship more clearly. For instance, in the exercise, \(\log_{36} 8 = a\) was rewritten in exponential form as \(36^a = 8\), helping us to work through the problem using familiar exponent rules.

Logarithm Properties

Logarithms have several notable properties that make them convenient for mathematical manipulation. Utilizing these properties can simplify complex logarithmic expressions, making them more manageable to solve or evaluate.
Some key properties include:

• Product Property: \(\log_b (mn) = \log_b m + \log_b n\)
• Quotient Property: \(\log_b \left(\frac{m}{n}\right) = \log_b m - \log_b n\)
• Power Property: \(\log_b (m^n) = n \log_b m\)

These properties simplify expressions by breaking down complex logarithms into simpler parts, often leading to easier calculations. For example, in our solution, the power property was applied when calculating \(\log_2 (3^2)\) and \(\log_2 (6^2)\), allowing the expression to be simplified to a form that could be more easily evaluated. By understanding and applying these properties, you can become more adept at solving logarithmic equations and manipulation.