Question

$\log_{6}16 = \frac{\log 16}{\log 6} = \frac{4\log 2}{\log 2 + \log 3}$

• A. $= \frac{4\log 2}{\log 2 + \frac{2a\log 2}{3 - a}} = \frac{4(3 - a)}{3 - a + 2a} = 4.\frac{3 - a}{3 + a}$
• B. $\log_{16}x = y \Rightarrow y^{2} - y + \log_{16}k = 0$
• C. $\therefore$
• D. $( - 1)^{2} - 4.1.\log_{16}k = 0 \Rightarrow 1 = \log_{16}k^{4}$

Answer 1

Answer: (D)

$( - 1)^{2} - 4.1.\log_{16}k = 0 \Rightarrow 1 = \log_{16}k^{4}$

Answer 2

$x = y$

$y = z$.