If $\log _2 x + \log _4 x + \log _8 x + \log _{16} x = \frac{25}{36}$ and $x = 2^k$,then $k$ is

If $x = \log_a (bc)$, $y = \log_b (ca)$, and $z = \log_c (ab)$, then the value of $\frac{1}{1+x} + \frac{1}{1+y} + \frac{1}{1+z}$ is

What is the number of real values of the parameter $k$ for which the equation $({\log _{16}}x)^2 - {\log _{16}}x + {\log _{16}}k = 0$ has exactly one solution,given that the coefficients are real?

If $x = \log_3 5$ and $y = \log_{17} 25$,then which of the following is correct?

The solution set of the equation $\left| {1 - {{\log }_{1/6}}x} \right| + \left| {{{\log }_2}x} \right| + 2 = \left| {3 - {{\log }_{1/6}}x + {{\log }_{1/2}}x} \right|$ is $\left[ {\frac{a}{b},a} \right]$,where $a, b \in N$. Then the value of $(a + b)$ is:

$x = \log \left( \frac{1}{y} + \sqrt{1 + \frac{1}{y^2}} \right) \Rightarrow y$ is equal to