document.write( "Question 407010: how do i solve (3y-1)^6=80 using logarithms? \n" ); document.write( "
Algebra.Com's Answer #287068 by lwsshak3(11628)\"\" \"About 
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how do i solve (3y-1)^6=80 using logarithms?\r
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\n" ); document.write( "\n" ); document.write( "I'm not sure logarithms is the best method to use, but you can easily solve it with regular algebra and a good calculator.\r
\n" ); document.write( "\n" ); document.write( "(3y-1)^6=80
\n" ); document.write( "raise each side to the (1/6) power
\n" ); document.write( "3y-1=80^(1/6)=2.076
\n" ); document.write( "3y=2.076+1=3.076
\n" ); document.write( "y=3.076/3=1.025\r
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\n" ); document.write( "\n" ); document.write( "Anyway, if you want to do it by logs, this is how you do it:\r
\n" ); document.write( "\n" ); document.write( "6log(3y-1)=log80
\n" ); document.write( "log(3y-1)=log80/6=.31718
\n" ); document.write( "change to exponential form,
\n" ); document.write( "3y-1=10^.31718=2.0758
\n" ); document.write( "3y=2.0758+1=3.0758
\n" ); document.write( "y=3.0758/3=1.025\r
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