If $\log 2=a, \log 3=b, \log 7=c$ and $6^x=7^{x+4}$, then $x$ is equal to:

Let $a, b, c$ be real numbers, each greater than $1$, such that $\frac{2}{3} \log _{b} a+\frac{3}{5} \log _{c} b+\frac{5}{2} \log _{a} c=3$. If the value of $b$ is $9$, then the value of $a$ must be

If $2 \log (x+1)-\log (x^{2}-1)=\log 2$, then $x=$

The solution to the equation ${\log _7}{\log _5}(\sqrt {{x^2} + 5 + x} ) = 0$ is:

If $x, y, z \in R^+$ are such that $z > y > x > 1$,$\log_{y}x + \log_{x}y = \frac{5}{2}$ and $\log_{z}y + \log_{y}z = \frac{10}{3}$,then $\log_{x}z$ is equal to

If ${\log _5}a \cdot {\log _a}x = 2$,then the value of $x$ is: