Question 173139
a chunk of radioactive material decays from a mass 
of 450 grams to 165 grams in two weeks,(14 days). 
<pre><font size = 4 color = "indigo"><b>
The exponential equation to use is:
{{{A= Pe^(r*t)}}}, where
{{{A}}} = Amount, in grams
{{{t}}} = time, in days
{{{r}}} = a constant
{{{P}}} = a constant

When {{{t = 0}}} days, {{{A = 450}}} grams, therefore
({{{t}}},{{{A}}}) = ({{{0}}},{{{450}}}) is a point on the graph.

When {{{t = 14}}} days, {{{A = 165}}} grams, therefore 
({{{t}}},{{{A}}}) = ({{{14}}},{{{165}}}) is another point on the graph.
 
Substitute the first point in the equation:

{{{A= Pe^(r*t)}}}

{{{450= Pe^(r*0))}}}

{{{450=Pe^0}}}

{{{450=P(1)}}}

{{{450 = P}}}

So we have found that P is the original amount
Substitute the second point in the equation:

{{{A= 450e^(r*t)}}}

{{{165= 450e^(r*14)}}}

{{{165= 450e^(14r)}}}

Divide both sides by 450:

{{{165/450= (450e^(14r)/450)}}}

{{{11/30 = e^(14r)}}}

Use the fact that {{{Y=e^X}}} is equivalent to {{{X=ln(Y)}}}
to rewrite the equation in natural log form:

{{{14r=ln(11/30)}}}

Divide both sides by 14

{{{r=ln(11/30)/14}}}

Get your calculator and find the right side:

{{{r = -.0716644363}}}

Substitute {{{-.0716644363}}} for {{{r}}} in

{{{A= 450e^(r*t)}}}

{{{A= 450e^(-.0716644363*t)}}}

Now we can do the last two parts:
</pre></font></b>
>>...find the half life time...<< 
<pre><font size = 4 color = "indigo"><b>
This asks us to find the time it takes
for the original 450 grams to reduce to
just half of that amount or 225 grams
(half of 450 grams)

So we substitute 225 for A in

{{{A= 450e^(-.0716644363*t)}}}

{{{225= 450e^(-.0716644363*t)}}}

Divide both sides by 450:

{{{225/450= (450e^(-.0716644363*t))/450}}}

{{{1/2 = e^(-.0716644363*t)}}}

Again use the fact that {{{Y=e^X}}} is equivalent to 
{{{X=ln(Y)}}} to rewrite the equation in natural 
log form:

{{{(-.0716644363*t)=ln(1/2)}}}

divide both sides by {{{-.0716644363}}}

{{{(-.0716644363*t)/(-.0716644363)=ln(1/2)/(-.0716644363)}}}

{{{t=ln(1/2)/(-.0716644365)}}}

Get your calculator and find the right side:

{{{t=9.672122128}}}

So, it takes about 9.7 days for the radioactive
material to reduce from its original amount
of 450 grams to half its original amount, or
225 grams.

Now the last part:
</pre></font></b>
>>...and the amount of material left after 70 days...<<
<pre><font size = 4 color = "indigo"><b>
All we have to do is substitute 70 for t in

{{{A= 450e^(-.0716644363*t)}}}

{{{A= 450e^(-.0716644363*70)}}}

{{{A = 450e^(-5.016510544)}}}

{{{A = 2.982425926}}}

So after 70 days, there is only about 3 grams
remaining.

Edwin</pre>