1.

Find the value of `root(3)(16.51)` approximately.

Answer» Let P = `root(3)(16.51)`
log P = log `(16.51)^(1//3)`
`=(1)/(3)log16.51`
`=(1)/(3) (1.2178)=0.4059`
log P = 0.4059
P = antilog (0.4059)
`therefore` P = 2.546.


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