# A particular compound decays according to the equation `y=ae^(-0.097t)`, where t is in days. Find the half-life of this compound.

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### 2 Answers

Half-life is the amount of time it takes for the compound to decay to the half the amount there was originally.

Originally, or at the moment with t = 0, the amount of the compound was

`y=ae^(-0.97*0) = a`

So, half of that is `y=a/2` .

To find time *t* when this happens, plug this value of *y *into the equation:

`a/2 = ae^(-0.97t)`

Divide both sides by *a*:

`1/2 = e^(-0.97t)`

`ln(1/2) = -0.97t`

`t=(ln(1/2))/(-0.97) = (ln(2))/0.97`

`t = 0.71 `

**The half-life of this compound is approximately 0.71 days.**

`y=ae^(0.097t)`

Half life: `t_h=1/2 y(0)=a/2`

`ae^(-0.097t_h)=a/2`

`e^(-0.097t_h)=1/2`

Using logaritms:

`-0,097t_h= ln(1/2)=-ln2`

`t_h= ln 2/0.097` `=7^d 3^h 30^m 01^s 20^c`

Let you see tree line of compund decays

Red curve line: a=50, black curve line: a=30, blue curve line : a=20

Relative colored straight linea stand for half life decays:

(Red straight line : y= 25, Black straight line: y=15, Blue straight line: y=10)

Note, as calculus shown, the intersection of half life with rispective curve of decays line, does'nt depend by a, indeed the point of intersection have the same value of `t_h`

`t_h=7.145847222267477416672496097507`

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