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avatar+2712 

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

tertre  Apr 14, 2017
 #1
avatar+86890 
+2

The first says that     4^x  =  3

 

The second says that  2^(kx)  = 27     

 

Re-writing the first, we have

 

 [ 2^2]^x  = 3

 

[2^x]^2  = 3    cube both sides

 

( [2^x]^2 )^3   = 27

 

2^(6x)  = 27

 

Which implies that  

 

2^(6x)  =   2^(kx)

 

Which implies that  k  = 6

 

 

cool cool cool 

CPhill  Apr 14, 2017
 #3
avatar+7085 
+2

Oh! This makes sense! laughlaughlaugh

I couldn't figure it out earlier! smiley

hectictar  Apr 14, 2017
edited by hectictar  Apr 14, 2017
 #2
avatar+2712 
+1

Thanks!

tertre  Apr 14, 2017

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