Problem 65 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { Given that } \log _{4} 125=a \text { , find } \log 64 \text { in terms of } a \text { . } $$

Step-by-Step Solution

Answer

The short answer is: \( \log{64} = \frac{3\log{125}}{a} \).

Step 1: Identify the given information

We are given the following information: \( \log_4{125} = a \) We are asked to find \( \log {64} \) in terms of 'a'.

Step 2: Change the base of the given logarithm

Recall the change of base formula: \( \log_b{c} = \frac{\log_d{c}}{\log_d{b}} \) Now, we can use the base 10 instead of base 4. Applying the change of base formula for the given information: \( a = \log_4{125} \) \( a = \frac{\log{125}}{\log{4}} \) Now we have the expression for 'a' in terms of base 10 logarithms.

Step 3: Express \(\log{64}\) in terms of 'a'

Now that we have 'a' expressed in base 10 logarithms, let's write the expression for \(\log{64}\) in terms of 'a'. We want to find \(\log{64}\). Using the relation we found in step 2: \( a = \frac{\log{125}}{\log{4}} \) Now we need to find an expression for \(\log{64}\) in terms of \(\log{125}\) and \(\log{4}\). We can write \(64\) as a power of \(4\): \( 64 = 4^3 \) Now we will take the logarithm of both sides: \( \log{(4^3)} = \log{64} \) Apply the logarithmic power rule to bring down the exponent: \( 3 \log{4} = \log{64} \) Since we know \( a = \frac{\log{125}}{\log{4}} \), we can write \(\log{4}\) in terms of 'a': \( \log{4} = \frac{\log{125}}{a} \) Now replace \(\log{4}\) in the expression for \(\log{64}\): \( \log{64} = 3 \cdot \frac{\log{125}}{a} \) So the expression for \(\log{64}\) in terms of a: \( \log{64} = \frac{3\log{125}}{a} \)

Key Concepts

Change of Base Formula

Understanding the change of base formula is essential when working with logarithms that are not a common base, such as base 10 or base 2. The formula allows you to convert a logarithm with any base to a different base to make calculations easier. The change of base formula is expressed as:

• \( \log_b{c} = \frac{\log_d{c}}{\log_d{b}} \)

Here, \( b \) is the original base while \( d \) is the new base, often chosen to be 10 for simplicity since calculators readily compute \( \log_{10} \). Changing to base 10, for example, transforms \( \log_4{125} \) into an expression using base 10 logs:

• \( a = \frac{\log{125}}{\log{4}} \)

This transformation is not just for theoretical purposes; it is practical for solving problems where you need to express logarithms in terms of known variables, like 'a' in this exercise.

Logarithmic Identities

Logarithmic identities are tools that simplify complex logarithmic expressions. One crucial identity used in this exercise is the power rule of logarithms, which states:

• \( \log{(c^n)} = n \cdot \log{c} \)

This property allows us to bring down exponents, which is particularly useful when working with numbers like 64 and 4 in the problem. By recognizing that \( 64 = 4^3 \), and applying the power rule, we express \( \log{64} \) as:

• \( \log{64} = 3 \cdot \log{4} \)

In this step, we take advantage of the rule to easily work with magnitudes and bases that simplify calculations or variable expressions.

Algebraic Manipulation

Algebraic manipulation in the context of logarithms is crucial to express one logarithm in terms of another. It involves reworking expressions to suit the terms you have, like changing \( \log{4} \) to be expressed in terms of 'a'. Since \( a = \frac{\log{125}}{\log{4}} \), we rearrange it to get \( \log{4} = \frac{\log{125}}{a} \). This helps replace terms and variables, allowing us to derive other values such as:

• \( \log{64} = 3 \cdot \frac{\log{125}}{a} \)

Through such rearrangements, you can leverage known information to calculate unknowns, facilitating problem-solving by transforming one set of mathematical expressions into another that's more suited to the conditions or information that you have.