If ${x^y} = {y^x}$,then ${(x/y)^{(x/y)}} = {x^{(x/y) - k}}$,where $k = $

Let $a, b, x$ be positive real numbers with $a \neq 1, x \neq 1, ab \neq 1$. Suppose $\log_{a} b = 10$,and $\frac{\log_{a} x \cdot \log_{x}(\frac{b}{a})}{\log_{x} b \cdot \log_{ab} x} = \frac{p}{q}$,where $p$ and $q$ are positive integers which are coprime. Then $p+q$ is

If $\log _k x \cdot \log _5 k = \log _x 5$,where $k \neq 1$ and $k > 0$,then the value of $x$ is:

The number of solutions of the equation $\log _{4}(x-1)=\log _{2}(x-3)$ is

If $a^x = b^y = c^z = d^w$,the value of $x\left(\frac{1}{y} + \frac{1}{z} + \frac{1}{w}\right)$ is

$7 \log \left( \frac{16}{15} \right) + 5 \log \left( \frac{25}{24} \right) + 3 \log \left( \frac{81}{80} \right)$ is equal to