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

 Apr 16, 2018
 #1
avatar+99520 
+3

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

 

 

cool cool cool

 Apr 16, 2018
 #2
avatar+21978 
+2

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

 

\(\left. \begin{array}{rcll} \log_{4}3 &=& x \\ \log_{2}27 &=& kx \\ \end{array} \right\} \begin{array}{|rcll|} \hline \dfrac{\log_{2}27}{\log_{4}3} &=& \dfrac{kx}{x} = k \qquad ( x\ne 0) \\ k &=& \dfrac{\log_{2}27}{\log_{4}3} \\\\ &=& \dfrac{ \dfrac{\ln(27)}{\ln(2)} }{ \dfrac{\ln(3)}{\ln(4)} } \\\\ &=& \dfrac{\ln(3^3)\ln(2^2)}{\ln(2)\ln(3)} \\\\ &=& \dfrac{3\ln(3)\cdot 2\ln(2)}{\ln(2)\ln(3)} \\\\ &=& 3\cdot 2 \\\\ \mathbf{k} & \mathbf{=} & \mathbf{6} \\ \hline \end{array}\)

 

laugh

 Apr 16, 2018

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