# The value of x satisfying the equation?

## $\frac{1}{\log} _ 3 \left(1994\right) + \frac{1}{\log} _ 6 \left(1994\right) + \frac{1}{\log} _ 9 \left(1994\right) + \frac{1}{\log} _ 12 \left(1994\right) = \frac{1}{{\log}_{3 x} \left(1994\right)}$

$\frac{1}{\log} _ 3 \left(1994\right) + \frac{1}{\log} _ 6 \left(1994\right) + \frac{1}{\log} _ 9 \left(1994\right) + \frac{1}{\log} _ 12 \left(1994\right) = \frac{1}{{\log}_{3 x} \left(1994\right)}$
$\implies {\log}_{1994} 3 + {\log}_{1994} 6 + {\log}_{1994} 9 + {\log}_{1994} 12 = {\log}_{1994} \left(3 x\right)$
$\implies {\log}_{1994} \left(3 \times 6 \times 9 \times 12\right) = {\log}_{1994} \left(3 x\right)$
$\implies 3 x = 3 \times 6 \times 9 \times 12$
$\implies x = 648$