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what would (3)log to the base 4/9 of (27/8)^(1/4)

 Aug 24, 2017
 #1
avatar+118608 
+2

Let

 

\(y=log_{(4/9)}(27/8)^{(3/4)}\\ y= \frac{log(27/8)^{(3/4)}}{log(4/9)}\\ y= \frac{(3/4)log(27/8)}{log(2/3)^2}\\ y= \frac{(3/4)log(3/2)^3}{log(2/3)^2}\\ y= \frac{(9/4)log(3/2)}{2log(2/3)}\\ y= \frac{9log(3/2)}{8log(2/3)}\\ y= \frac{9log(3/2)}{8log(3/2)^{-1}}\\ y= \frac{9log(3/2)}{-8log(3/2)}\\ y=\frac{-9}{8} \)

 Aug 25, 2017
 #2
avatar+118608 
+2

OR

 

\(log_{(4/9)}(\frac{27}{8})^{3/4}\\ =log_{(4/9)}(\frac{3}{2})^{9/4}\\ =log_{(4/9)}(\frac{2}{3})^{-9/4}\\ =log_{(4/9)}(\frac{4}{9})^{-9/8}\\ =\frac{-9}{8}\)

 Aug 25, 2017

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