Logarithm Finding, solving for k.

Logarithm Finding, solving for k.

\(\text{Let $\log_{4}3=x$. Then $\log_{2}27=kx$. Find $k$.}\)

\(\large{ \begin{array}{|llcll|} \hline (1) & \log_{4}3=x & \text{or} & 4^x = 3 \\ (2) & \log_{2}27=kx & \text{or} & 2^{(kx)} = 27 \\ \hline \end{array} } \)

\(\large{ \begin{array}{|rcll|} \hline 2^{(kx)} &=& 27 \\ 2^{(\frac22 kx)} &=& 27 \\ 2^{(2\frac{kx}{2})} &=& 27 \\ \left(2^2\right)^{ \frac{kx}{2} } &=& 27 \\ 4^{( \frac{kx}{2} )} &=& 27 \\ 4^{( x\frac{k}{2} )} &=& 27 \\ \left(4^x\right)^{(\frac{k}{2} )} &=& 27 \quad & | \quad 4^x = 3 \\ 3^{(\frac{k}{2} )} &=& 27 \quad & | \quad 27 = 3^3 \\ 3^{(\frac{k}{2} )} &=& 3^3 \\ \frac{k}{2} &=& 3 \\ \mathbf{k} & \mathbf{=} & \mathbf{6} \\ \hline \end{array} } \)

Hello, I'm back! 

\(\log_{4}3=x\), 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 \(\boxed{k=6}\) . 

Solve for k.
2^((log(3) k)/log(4)) = 27

Take the logarithm base 2 of both sides:
(log(3) k)/log(4) = log(27)/log(2)

Divide both sides by log(3)/log(4):

k = (log(4) log(27))/(log(2) log(3))= 6