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| Question 407010:  how do i solve (3y-1)^6=80 using logarithms?
 Answer by lwsshak3(11628)
      (Show Source): 
You can put this solution on YOUR website! how do i solve (3y-1)^6=80 using logarithms? ..
 I'm not sure logarithms is the best method to use, but you can easily solve it with regular algebra and a good calculator.
 (3y-1)^6=80
 raise each side to the (1/6) power
 3y-1=80^(1/6)=2.076
 3y=2.076+1=3.076
 y=3.076/3=1.025
 ..
 Anyway, if you want to do it by logs, this is how you do it:
 6log(3y-1)=log80
 log(3y-1)=log80/6=.31718
 change to exponential form,
 3y-1=10^.31718=2.0758
 3y=2.0758+1=3.0758
 y=3.0758/3=1.025
 
 
 
 
 
 
 
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