+26376 Logarithm Finding, solving for k.
\(\text{Let $\log_{4}3=x$. Then $\log_{2}27=kx$. Find $k$.}\)
\(\large{ \begin{array}{|llcll|} \hline (1) & \log_{4}3=x & \text{or} & 4^x = 3 \\ (2) & \log_{2}27=kx & \text{or} & 2^{(kx)} = 27 \\ \hline \end{array} } \)
\(\large{ \begin{array}{|rcll|} \hline 2^{(kx)} &=& 27 \\ 2^{(\frac22 kx)} &=& 27 \\ 2^{(2\frac{kx}{2})} &=& 27 \\ \left(2^2\right)^{ \frac{kx}{2} } &=& 27 \\ 4^{( \frac{kx}{2} )} &=& 27 \\ 4^{( x\frac{k}{2} )} &=& 27 \\ \left(4^x\right)^{(\frac{k}{2} )} &=& 27 \quad & | \quad 4^x = 3 \\ 3^{(\frac{k}{2} )} &=& 27 \quad & | \quad 27 = 3^3 \\ 3^{(\frac{k}{2} )} &=& 3^3 \\ \frac{k}{2} &=& 3 \\ \mathbf{k} & \mathbf{=} & \mathbf{6} \\ \hline \end{array} } \)
+4614 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 \(\boxed{k=6}\) .
