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Logarithm Finding, solving for k.

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

Aug 21, 2018

#1
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Logarithm Finding, solving for k.

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

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

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

Aug 21, 2018
#2
+4249
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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 $$\boxed{k=6}$$

Aug 21, 2018
#3
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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

Aug 21, 2018