Question 1070297
<pre>
{{{A=Pe^(RT)}}}

where 
      P = the beginning amount
      T = the number of years
      R = the rate of decay
      A = amount left after T years.

In the first case we are given that
when T = 200000000=2*10^8, A = (80% of P) = 0.8P

So we substitute 200000000 for T and 0.8P for A

{{{0.8P=Pe^(R(200000000))}}}

P's cancel on both sides:

{{{0.8=e^(R(200000000))}}}

Take natural logarithms of both sides:

{{{ln(0.8)=ln(e^(R(200000000)))}}}

{{{ln(0.8)= R(200000000)}}}

{{{ln(0.8)/(200000000)=R}}}

Do that on a calculator and get 

{{{-1.11571776*10^(-9)= R}}}

So now the formula 

{{{A=Pe^(RT)}}}

becomes

{{{matrix(2,1,"",A=Pe^((-1.11571776*10^(-9))T))}}}

And since we want to know what percentage of the
original amount P, we let x be the decimal equivalent 
that we must multiply the original amount P by, so
we substitute xP for A and T=600000000 = 6x10<sup>8</sup>

{{{matrix(2,1,"",xP=Pe^((-1.11571776*10^(-9))600000000))}}}

Cancel the P's and add simplify the exponent

{{{matrix(2,1,"",x=e^((-6.69430656*10^(-1))))}}}
  
{{{matrix(2,1,"",x=e^(-0.669430656))}}}

{{{matrix(2,1,"",x=0.5119999989)}}}

That rounds to 0.51 or 51%

--------------------------------------

The half-life is when A = one-half of the original amount P

So we substitute {{{expr(1/2)P}}} for A

{{{matrix(2,1,"",A=Pe^((-1.11571776*10^(-9))T))}}}

{{{matrix(2,1,"",expr(1/2)P=Pe^((-1.11571776*10^(-9))T))}}}

The P's cancel

{{{matrix(2,1,"",expr(1/2)=e^((-1.11571776*10^(-9))T))}}}

Take natural logs of both sides:

{{{matrix(2,1,"",ln(1/2)=ln(e^((-1.11571776*10^(-9))T)))}}}  

{{{matrix(2,1,"",ln(1/2)=(-1.11571776*10^(-9))T)}}}

{{{matrix(2,1,"",ln(1/2)/(-1.11571776*10^(-9))=T)}}}

Do that on a calculator and get

{{{matrix(2,1,"",621256742=T)}}}

which rounds to 621000000 or 621 million years.

[Note: this would be easier if you would use the STO
key on your TI calculator to store the messy values,
so you could just use the letters. Maybe you just
round off as you go.]

Edwin</pre>