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December 20, 2014

December 20, 2014

Posted by **Katie** on Tuesday, March 5, 2013 at 10:59pm.

- math -
**Steve**, Tuesday, March 5, 2013 at 11:19pmassuming logs base 10,

2logx = log_6(8)

logx = 1/2 log_6(8)

x = √(10^(log_6(8))

now log_6(8) = log6/log8. so

x = √(10^log6)^(1/log8)

= √(6^1/log8)

= 6^(1/log64)

- math -
**Reiny**, Tuesday, March 5, 2013 at 11:19pmtake log of both sides

log [ 6^(2logx) ] = log8

2logx (log6) = log8

2logx = log8/log6

log x^2= log8/log6

x^2= 10^(log8/log6)

x = √( 10^(log8/log6) ) = 3.804339012

check using your calculator, it works

- math - go with Reiny -
**Steve**, Wednesday, March 6, 2013 at 3:34pmGo with Reiny's answer. I messed up my log base conversion.

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