Question 191254
<pre><font size = 4 color="indigo"><b> 
{{{A = Pe^rt}}}

When {{{t=0}}}, {{{A=18}}}, plug those in:

{{{A = Pe^rt}}}
{{{18 = Pe^r(0)}}}
{{{18 = Pe^0}}}
{{{18 = P(1)}}}
{{{18 = P}}}

Substitute 18 for P in

{{{A = 18e^rt}}}

When {{{t=3}}}, {{{A=6}}}, plug those in:

{{{A = 18e^rt}}}
{{{6 = 18e^r(3)}}}
{{{6 = 18e^(3r)}}}

Divide both sides by 18

{{{ 6/18 = ( 18*e^(3r) )/18 }}}
{{{1/3 = e^(3r)}}}

Use principle {{{A=e^N}}} is equivalent to {{{N=ln(A)}}}

{{{3r=ln(1/3)}}}

Divide both sides by 3

{{{r = ln(1/3)/3}}}

In your TI-84 type  ln(1/3)/3 ENTER and get -.3662040962

{{{r = -.3662040962 }}}

Substitute -.3662040962 for r in

{{{A = 18e^(rt)}}}

{{{A = 18e^(-.3662040962t)}}} 

Now we substitute 11 for t

{{{A = 18e^(-.3662040962t)}}} 
{{{A = 18e^(-.3662040962(11))}}}
{{{A = 18e^(-4.028245058)}}}
{{{A = 18(.0178055503)}}}
{{{A = .3204999045}}}

So after 11 years, there are only .3205 grams.

Edwin</pre>