Question

Suppose $a, b, c$ are positive real numbers such that $2^a = 3^b = 6^{-c}$. Then $\frac{ab + bc + ca}{abc} =$ ?

• A. 0
• B. 1
• C. 6
• D. 3

Hint:

Use $\ln(ab)=\ln a+\ln b$ to simplify logarithmic expressions quickly.

Answer 1

The Correct Option is A

Concept: Take common value and convert into logarithmic form.
Step 1: Let common value = $k$.
\[ 2^a = 3^b = 6^{-c} = k \]
Step 2: Take logarithm.
\[ a\ln2 = \ln k,\quad b\ln3 = \ln k,\quad -c\ln6 = \ln k \]
Step 3: Express variables.
\[ a = \frac{\ln k}{\ln2},\quad b = \frac{\ln k}{\ln3},\quad c = -\frac{\ln k}{\ln6} \]
Step 4: Substitute in expression.
\[ \frac{ab + bc + ca}{abc} = \frac{1}{c} + \frac{1}{a} + \frac{1}{b} \]
Step 5: Evaluate.
\[ \frac{1}{a} = \frac{\ln2}{\ln k},\quad \frac{1}{b} = \frac{\ln3}{\ln k},\quad \frac{1}{c} = -\frac{\ln6}{\ln k} \] \[ \Rightarrow \frac{\ln2 + \ln3 - \ln6}{\ln k} \] \[ = \frac{\ln(2 \cdot 3) - \ln6}{\ln k} = \frac{\ln6 - \ln6}{\ln k} = 0 \]
Hence, the value is 0.