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

Answer 1

Let: \[ 14^a = 36^b = 84^c = k \] Then taking logarithms: \[ a = \log_{14} k \Rightarrow \frac{1}{a} = \log_k 14 \] Similarly: \[ b = \log_{36} k, \quad \frac{1}{c} = \log_k 84 \]

We are asked to find the value of the expression: \[ 6b \left( \frac{1}{c} - \frac{1}{a} \right) \]

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

Using log subtraction rule: \[ = 6 \cdot \log_{36} k \cdot \log_k \left( \frac{84}{14} \right) = 6 \cdot \log_{36} k \cdot \log_k 6 \]

Applying change of base: \[ \log_{36} k = \frac{1}{\log_k 36}, \quad \log_k 6 = \log_k 6 \] So: \[ = 6 \cdot \frac{1}{\log_k 36} \cdot \log_k 6 = 6 \cdot \frac{\log_k 6}{\log_k 36} \] Now note: \[ \log_k 36 = \log_k (6^2) = 2 \log_k 6 \Rightarrow \frac{\log_k 6}{2 \log_k 6} = \frac{1}{2} \]

Final expression: \[ 6 \cdot \frac{1}{2} = \boxed{3} \]

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Final Answer:

\(\boxed{3}\)

Answer 2

Given: \[ 14^a = 36^b = 84^c \] First, express all numbers in their prime factorizations: \[ 14 = 2 \cdot 7, \quad 36 = 2^2 \cdot 3^2, \quad 84 = 2^2 \cdot 3 \cdot 7 \]

Raising each to the powers \( a \), \( b \), and \( c \) respectively: \[ 14^a = 2^a \cdot 7^a,\quad 36^b = 2^{2b} \cdot 3^{2b},\quad 84^c = 2^{2c} \cdot 3^c \cdot 7^c \]

Since all three expressions are equal: \[ 2^a \cdot 7^a = 2^{2b} \cdot 3^{2b} = 2^{2c} \cdot 3^c \cdot 7^c \] Comparing powers of each prime:

• \(2^a = 2^{2b} \Rightarrow a = 2b\)
• \(2^a = 2^{2c} \Rightarrow a = 2c\)

Hence: \[ a = 2b = 2c \Rightarrow b = c \]

 

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

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Final Answer:

\(\boxed{3}\)