Let $\left(x_0, y_0\right)$ be the solution of the following equations $(2 x)^{\ln 2} =(3 y)^{\ln 3}$ $3^{\ln x} =2^{\ln y}$ . Then $x_0$ is
$\frac{1}{6}$
$\frac{1}{3}$
$\frac{1}{2}$
$6$
The value of $(0.16)^{\log _{2.5}\left(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots . to \infty\right)}$ is equal to
If ${1 \over {{{\log }_3}\pi }} + {1 \over {{{\log }_4}\pi }} > x,$ then $x$ be
If $n = 1983!$, then the value of expression $\frac{1}{{{{\log }_2}n}} + \frac{1}{{{{\log }_3}n}} + \frac{1}{{{{\log }_4}n}} + ....... + \frac{1}{{{{\log }_{1983}}n}}$ is equal to
${\log _4}18$ is
Let $a , b , c$ be three distinct positive real numbers such that $(2 a)^{\log _{\varepsilon} a}=(b c)^{\log _e b}$ and $b^{\log _e 2}=a^{\log _e c}$. Then $6 a+5 b c$ is equal to $........$.