Problem 23 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

\(f(x)=\log _{0.5}\left\\{-\log _{2}\left(\frac{3 x-1}{3 x+2}\right)\right\\}\)

Step-by-Step Solution

Answer

The simplified function is \(f(x)= \log _{2}\left(\frac{3 x-1}{3 x+2}\right)\).

Step 1: Change the base for the outer logarithm

The given function has an outer logarithm with base 0.5. However, it is easier to work with logarithms of base 2. To do that, we use the change of base formula: \[\log_a b = \frac{\log_c b}{\log_c a}\] Applying this formula to our function, we get: \[f(x)=\frac{\log_2\left\\{-\log _{2}\left(\frac{3 x-1}{3 x+2}\right)\right\\}}{\log_2 0.5}\]

Step 2: Simplify the denominator

Observe that the denominator of our fraction is now a logarithm with base 2 of 0.5. We know that \(2^{-1} = 0.5\), so using properties of logarithms: \[\log_2 0.5 = \log_2 2^{-1} = -1\] Now, substitute this value back into the function: \[f(x)=\frac{\log_2\left\\{-\log _{2}\left(\frac{3 x-1}{3 x+2}\right)\right\\}}{-1}\]

Step 3: Simplify the function further

Since we are dividing by -1, we can instead multiply by -1: \[f(x)=-\log_2\left\\{-\log _{2}\left(\frac{3 x-1}{3 x+2}\right)\right\\}\] Now, we need to get rid of the double logarithm, which we can do using the property \(\log_a(b^c) = c\log_a(b)\). In this case, we have: \[f(x)= -\log _{2}\left(2^{-\log _{2}\left(\frac{3 x-1}{3 x+2}\right)}\right)\]

Step 4: Simplify using properties of logarithms

Now, using the property \(\log_a(b^c) = c\log_a(b)\) again: \[f(x)= -\left(-\log _{2}\left(\frac{3 x-1}{3 x+2}\right)\right)\] The two negative signs in the expression cancel each other out: \[f(x)= \log _{2}\left(\frac{3 x-1}{3 x+2}\right)\] Our simplified function is:

Step 5: Final Solution

\[f(x)= \log _{2}\left(\frac{3 x-1}{3 x+2}\right)\]

Key Concepts

Change of Base Formula

Changing the base of a logarithm is a helpful trick that simplifies complex expressions. When dealing with different bases, the Change of Base Formula becomes valuable:

• \( \log_a b = \frac{\log_c b}{\log_c a} \)

This formula allows you to convert a logarithm from one base to another, typically to a base more convenient for calculations, such as base 10 or base 2.
To apply this, you take the logarithm of the number and divide it by the logarithm of the original base, both using the new base.
For example, in the given function: \[ f(x) = \log_{0.5} \{-\log_{2} \frac{3x-1}{3x+2}\} \]We chose to convert the outer logarithm to a base of 2 using this formula:\[ f(x) = \frac{\log_2 \{-\log_{2} \left( \frac{3x-1}{3x+2} \right)\}}{\log_2 0.5} \] This sets the stage for easier simplification in later steps.

Properties of Logarithms

Properties of logarithms provide essential tools for algebraically manipulating and simplifying expressions. Some important properties to remember 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: \( \log_b (M^k) = k \log_b M \)
• Base Change Property: This is used to convert log bases using the Change of Base Formula.

In our exercise, the Power Property was prominent:\[ \log_a (b^c) = c \log_a (b) \]This property facilitated the simplification of the expression by addressing exponents within the logarithmic argument.
As seen, this particular property allowed us to simplify the double logarithm in the function to achieve:\[ f(x)= \log _{2}\left(\frac{3 x-1}{3 x+2}\right) \]
Effectively, using these properties helps make complex logarithmic expressions more manageable.

Simplification of Logarithmic Expressions

Simplifying logarithmic expressions is the ultimate goal to make functions easier to understand and work with.This involves reducing expressions to their most compact form, often using a mix of strategies from algebra and logarithm properties.
In the original function:\[ f(x) = \log_{0.5} \{-\log_{2} \left( \frac{3x-1}{3x+2} \right) \} \]The first step was applying the Change of Base Formula, transforming the base to 2, a more straightforward option:\[ f(x) = \frac{\log_2 \{-\log_{2} \left( \frac{3x-1}{3x+2} \right)\}}{\log_2 0.5} \]Then, we evaluated the logarithm in the denominator \( \log_2 0.5 \), simplifying it to -1 since \( 2^{-1} = 0.5 \).
Finally, by leveraging the Power Property and cancelling out negatives, the function resolved to a simpler form:\[ f(x)= \log_{2}\left(\frac{3x-1}{3x+2}\right) \]
This illustrates how step-by-step simplification transforms a cumbersome expression into a concise result, enhancing both understanding and utility.