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Evaluate `root(7)(0.00003587)`. |
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Answer» 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. |
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