Question

$B = 0,C = - 2$

• A. 1
• B. 7/13
• C. 13/7
• D. None of these

Answer 1

Answer: (B)

7/13

Answer 2

Let $\log_{49}28 = \frac{\log 28}{\log 49} = \frac{\log 7 + \log 4}{2\log 7}$, then $= \frac{\log 7}{2\log 7} + \frac{\log 4}{2\log 7} = \frac{1}{2} + \frac{1}{2}\log_{7}4$

$= \frac{1}{2} + \frac{1}{2}.2\log_{7}2 = \frac{1}{2} + \log_{7}2 = \frac{1}{2} + m = \frac{1 + 2m}{2}$, then $\log_{e}\left( \frac{a + b}{2} \right) = \frac{1}{2}(\log_{e}a + \log_{e}b)$

$= \frac{1}{2}\log_{e}(ab) = \log_{e}\sqrt{ab}$ Given expression

= $\Rightarrow \frac{a + b}{2} = \sqrt{ab} \Rightarrow a + b = 2\sqrt{ab}$

$\Rightarrow (\sqrt{a} - \sqrt{b})^{2} = 0 \Rightarrow \sqrt{a} - \sqrt{b} = 0 \Rightarrow a = b.$

$\log_{10}\alpha,\log_{3}\alpha,\log_{e}\alpha,\log_{2}\alpha$

$\sqrt{\log_{0.5}^{2}4} = \sqrt{\{\log_{0.5}(0.5)^{- 2}\}^{2}} = \sqrt{( - 2)^{2}} = 2$

$\log_{3}4.\log_{4}5.\log_{5}6.\log_{6}7.\log_{7}8.\log_{8}9$

$= \frac{\log 4}{\log 3}.\frac{\log 5}{\log 4}.\frac{\log 6}{\log 5}.\frac{\log 7}{\log 6}.\frac{\log 8}{\log 7}.\frac{\log 9}{\log 8} = \frac{\log 9}{\log 3} = \log_{3}9 = \log_{3}3^{2} = 2$.