Let `x = root(7)(0.00003587).` Then,
log `x = 1/7` log (0.00003587)
`= 1/7 (bar(5).5548) = 1/7 (-5 + 0.5548)`
`= 1/7 (-4.4452) = -0.6350`
` = -1 + (1 - 0.6350) = bar(1).3650`
`rArr` x = antilog `(bar(1).3650) = 0.2317.`
Hence, the required value is 0.2317.