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# Help please (Happy birthday Melody!)

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

Apr 16, 2018

#1
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$$\log_{4}3=x. Then \log_{2}27=kx$$       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

Apr 16, 2018
#2
+24983
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Let $\log_{4}3=x$. Then $\log_{2}27=kx$. Find $k$.

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

Apr 16, 2018