Question

If \(a, b, c\) are non-zero and \(14^a = 36^b = 84^c\), then \(6b \bigg(\frac{1}{c}-\frac{1}{a}\bigg)\)is equal to [This Question was asked as TITA]

• A. 9
• B. 3
• C. 2
• D. 4

Answer 1

The Correct Option is B

Let \(14^a = 36^b = 84^c = k\) 

Taking logarithm on both sides: 
\(\Rightarrow a = \log_{14} k \Rightarrow \frac{1}{a} = \log_k 14\)
Similarly, \(\frac{1}{c} = \log_k 84\) and \(b = \log_{36} k\)

We are asked to evaluate: 
 

\[6b\left( \frac{1}{c} - \frac{1}{a} \right)\]

 Substituting the expressions, we get: 
 

\[6 \cdot \log_{36} k \cdot (\log_k 84 - \log_k 14)\]

Using the property \(\log_b a = \frac{1}{\log_a b}\), we can write: 
 

\[6 \cdot \frac{\log k}{\log 36} \cdot \left( \frac{\log 84}{\log k} - \frac{\log 14}{\log k} \right)\]

Simplifying: 

\[6 \cdot \frac{\log k}{\log 36} \cdot \frac{\log 84 - \log 14}{\log k}\]

 
 

\[= 6 \cdot \frac{\log 84 - \log 14}{\log 36}\]

 
 

\[= 6 \cdot \frac{\log \left( \frac{84}{14} \right)}{\log 36} = 6 \cdot \frac{\log 6}{\log 36}\]

Since \(\log 36 = \log 6^2 = 2 \log 6\), we get: 
 

\[6 \cdot \frac{\log 6}{2 \log 6} = 6 \cdot \frac{1}{2} = 3\]

Final Answer: \( \boxed{3} \)

Answer 2

We are given the equation:

\[ 14^a = 36^b = 84^c \]

First, express each base in terms of prime factors:

• \(14 = 2 \times 7 \Rightarrow 14^a = 2^a \times 7^a\)
• \(36 = 2^2 \times 3^2 \Rightarrow 36^b = 2^{2b} \times 3^{2b}\)
• \(84 = 2^2 \times 3 \times 7 \Rightarrow 84^c = 2^{2c} \times 3^c \times 7^c\)

So the common value becomes:

\[ 2^a \cdot 7^a = 2^{2b} \cdot 3^{2b} = 2^{2c} \cdot 3^c \cdot 7^c \]

By comparing exponents of matching prime bases, we get:

\[ a = 2b = 2c \Rightarrow b = \frac{a}{2},\quad c = \frac{a}{2} \]

Now calculate the expression:

\[ 6b \left( \frac{1}{c} - \frac{1}{a} \right) \] Substituting \(c = \frac{a}{2}\) and \(b = \frac{a}{2}\): \[ = 6 \cdot \frac{a}{2} \left( \frac{2}{a} - \frac{1}{a} \right) = 3a \cdot \frac{1}{a} = \boxed{3} \]

Final Answer:
\[ \boxed{3} \]