Problem 73 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { Prove that } \log _{a b} x=\frac{\log _{a} x \log _{b} x}{\log _{a} x+\log _{b} x} $$

Step-by-Step Solution

Answer

The given equation is proven by applying logarithm properties. The left hand side was simplified using the change of base formula and product rule, while the right hand side was manipulated by factoring. These changes allowed the two sides to match, thus proving the equation true.

Step 1: Apply the Change of Base Formula

The change of base formula states: \( \log_b a = \frac{\log_c a}{\log_c b} \). Applying this to \( \log_{ab} x \) we get \( \frac{\log_a x}{\log_a ab} \). Also apply it to \( \log_a x \) and \( \log_b x \) in the denominator of the right hand side to match the base.

Step 2: Simplify \( \log_a ab \)

Using the product rule of logs, which says \( \log_b mn = \log_b m + \log_b n \), simplify \( \log_a ab \) to \( 1 + \log_a b \). The left hand side becomes \( \frac{\log_a x}{1 + \log_a b} \).

Step 3: Manipulate The Right Hand Side

To confirm that the left hand side equals to the right hand side, it is needed to manipulate the right hand side to match this same format. The right hand side is \( \frac{\log_a x \log_b x}{\log_a x + \log_b x} \). Factor \( \log_a x \) out of the numerator and denominator. This changes the equation to \( \frac{\log_a x (\log_b x)}{\log_a x (1 + \log_b a)} \).

Step 4: Comparison and Conclusion

On comparing, the simplified version of the left hand side, \( \frac{\log_a x}{1 + \log_a b} \), matches exactly with the simplified version of the right hand side, \( \frac{\log_a x (\log_b x)}{\log_a x (1 + \log_b a)} \). This verifies the equality of the two, and so the equation is proven true.

Key Concepts

Change of Base Formula

The Change of Base Formula is a really handy tool when dealing with logarithms. Imagine you want to convert a logarithm that is not in base 10 (the common logarithm) or base e (the natural logarithm) to something easier to compute or further manipulate.
The formula is simple:

• Given as: \( \log_b a = \frac{\log_c a}{\log_c b} \).
• This means you can choose any base \( c \) that makes things easier, often 10 or \( e \).

Using this formula allows you to transform complex logarithms into terms you might be more equipped to handle. For example, \( \log_{ab} x \) can be rewritten as \( \frac{\log_a x}{\log_a ab} \).
This switch is an essential step in many logarithmic identities and proofs, making calculations more manageable.

Product Rule of Logarithms

The Product Rule of Logarithms is a fundamental property that simplifies logarithms of products. It states:

• \( \log_b (mn) = \log_b m + \log_b n \).

This rule reflects how exponents in multiplication behave, where you add the exponents for a common base. This rule becomes very useful for breaking down complex logs into simpler parts.
Consider \( \log_a ab \), using the product rule we have:

• \( \log_a ab = \log_a a + \log_a b = 1 + \log_a b \).

This simplification is crucial in many algebraic transformations involving logs, providing a more tangible form that is easier to handle in proofs and calculations.

Logarithmic Manipulation

Logarithmic manipulation is the art of using properties and identities of logarithms to transform and simplify expressions. It often involves applying rules like the Change of Base, Product Rule, Quotient Rule, and Power Rule to reframe the equation effectively.
In proving \( \log_{ab} x = \frac{\log_a x \log_b x}{\log_a x + \log_b x} \), manipulation is key:

• Start by applying the change of base to the left side.
• Use the product rule to simplify involved logs like \( \log_a ab \).
• Break down the right side, factoring out common terms \( \log_a x \).

Such smart manipulation helps to identify equal expressions, solve equations, and understand deeper connections in logarithmic relationships.