What is the ratio of `t_(1//2)` to `t_(1//3)` (for the amount of substance left) for first order reaction ?

Explanation

`a =` Initial amount
`(a-x) =` Amount left
`t_(1//2) = (2.3)/(k)log.((1)/(1//2)) = (2.3)/(k) log2` …(i)
`t_(1//3) = (2.3)/(k) log.(a)/(a-x)`
`= (2.3)/(k) log.(1)/(1//3)`
`= (2.3)/(k)log3` …(ii)
`:. (t_(1//2))/(t_(1//3)) = ((2.3//k)log2)/((2.3//k)log3) = (0.3)/(0.48) = 0.625`
Alternatively:
`(t_(1//2))/(t_(1//3)("left")) = (0.3)/(log.((1)/(1//3))) = (0.3)/(log 3) = (0.3)/(0.48) = 0.625`