1.

The activity of a certain preparation decreases 2.5 times after 7.0 days. Find its half time.

Answer»

SOLUTION :If the half-life is `T` days
`(2)^(-7//T)=(1)/(2.5)`
Hence `(7)/(T)= (In 2.5)/(In 2)`
or `T=(7In 2)/(In 2.5)= 5.30 days`


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