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The value of $\log_{a} \left( \frac {a}{b} \right) + \log_{b} \left( \frac{b}{a} \right),$ for $ 1 < a \leq b$ cannot be equal to 

  1. $ – 0.5$
  2. $1$
  3. $0$
  4. $ – 1$
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Given that, $1 < a \leq b$

Now, $ \log_{a} \left(\frac{a}{b} \right) + \log_{b} \left(\frac{b}{a} \right) = \log_{a}a – \log_{a}b + \log_{b}b – \log_{b}a $

$ \Rightarrow  \log_{a} \left(\frac{a}{b} \right) + \log_{b} \left(\frac{b}{a} \right) = 1 – \log_{a}b + 1 – \log_{b}a $

$\Rightarrow  \log_{a} \left(\frac{a}{b} \right) + \log_{b} \left(\frac{b}{a} \right) = 2 – \left( \log_{a}b + \log_{b}a \right) $

$\Rightarrow  \log_{a} \left(\frac{a}{b} \right) + \log_{b} \left(\frac{b}{a} \right)   = 2 – \left( \log_{a}b +\frac{1}{\log_{a}b} \right) \quad \longrightarrow (1) $

Now, using $\text{AM} \geq \text{GM}$ 

$ \frac{ \log_{a}b + \frac{1}{\log_{a}b}} {2} \geq \sqrt{ \log_{a}b \cdot \frac{1}{\log_{a}b}} $

$ \Rightarrow \log_{a}b + \frac{1}{\log_{a}b} \geq 2 $

From equation $(1),$

$ \log_{a} \left( \frac{a}{b} \right) + \log_{b} \left( \frac{b}{a} \right) = \underbrace{2 – \underbrace{\left( \log_{a}b + \frac{1}{\log_{a}b} \right)}_{\geq 2}}_{\leq 0}$

Thus, it can never be equal to $1.$

Correct Answer$: \text{B}$


$\textbf{PS:}$ Important logarithm properties : 

  • $\log_{a}a = 1$
  • $\log_{a} \left( \frac{m}{n} \right) = \log_{a}m – \log_{a}n $
  • $ \log_{a} (mn) = \log_{a}m + \log_{a}n $
  • $ \log_{a^{x}}b = \frac{1}{x} \log_{a}b $
  • $ \log_{a}b = \frac{1}{\log_{b}a} $
  • $ \log_{a}b = x \Rightarrow b = a^{x} $
  • $ \log_{a}b = \frac{\log_{c}b}{\log_{c}a} $
  • $ \log_{a}1 = 0 $
  • $ \log_{b}a^{x} = x \log_{b}a $
  • $ a^{\log_{a}b} = b $
  • $ \log_{a}^{k} n = \left( \log_{a}n \right)^{k} $
  • $ a^{\log_{b}{x}} = x^{\log_{b}{a}} $
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