+128406 \(\log_{4}3=x$. Then $\log_{2}27=kx\) find k
In exponential form we have.....
4^x = 3 and 2^(kx) = 27
Working with the expression on the left :
4^x = 3 and we can write
(2^2)^x = 3
2^(2x) = 3
Cube each side
[ 2^(2x)]^3 = 3^3
2^(6x) = 27
So k = 6
+26367 Let $\log_{4}3=x$. Then $\log_{2}27=kx$. Find $k$.
\(\left. \begin{array}{rcll} \log_{4}3 &=& x \\ \log_{2}27 &=& kx \\ \end{array} \right\} \begin{array}{|rcll|} \hline \dfrac{\log_{2}27}{\log_{4}3} &=& \dfrac{kx}{x} = k \qquad ( x\ne 0) \\ k &=& \dfrac{\log_{2}27}{\log_{4}3} \\\\ &=& \dfrac{ \dfrac{\ln(27)}{\ln(2)} }{ \dfrac{\ln(3)}{\ln(4)} } \\\\ &=& \dfrac{\ln(3^3)\ln(2^2)}{\ln(2)\ln(3)} \\\\ &=& \dfrac{3\ln(3)\cdot 2\ln(2)}{\ln(2)\ln(3)} \\\\ &=& 3\cdot 2 \\\\ \mathbf{k} & \mathbf{=} & \mathbf{6} \\ \hline \end{array}\)
