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Evaluate `root(3)(0.08034).`

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Evaluate `root(3)(0.08034).`
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Let `x = root(3)(0.08034).` Then,
log `x = 1/3` log (0.08034).
`= 1/3 (bar(2).9049) = 1/3(-2 + 0.9049) = 1/3(-1.0951)`
`= -0.3650 = -1 + (1 - 0.3650) = bar(1).6350`
`rArr` x = antilog `(bar(1).6350) = 0.4315.`
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