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\(\log_{4}3=x\), and \(\log_{2}27=kx\) .
This means, \(4^x=3\)\(2^{(kx)}=27\).
This can be broken up into: \((2^2)^x=3\) , and
\(2^{kx}=27\).
So, \(2^{2x}=3\), \(2^{kx}=27\).
Cubing the first equation, we have: \((2^{2x})^3=2^{6x}\) , and \(2^{kx}=27\).
By scanning, we can easily see that \(\boxed{k=6}\) .