Question 1201501
<pre>
At time=0, there are 6,000 grams of a radioactive material present. The half-life of the element is 18 years. In how many years will there be 115 grams remaining? Round your answer to the nearest 0.01 years.

Any help would be appreciated, I have tried many problems like this one without success. 

If ½-life is “a” time-periods, then k, or DECAY CONSTANT = {{{matrix(1,7, ln(1/2)/a, or, ln(.5)/a, "=", ln(.5)/18, "=", - .0385)}}}

CONTINUOUS GROWTH/DECAY formula: {{{matrix(1,3, A, "=", A[o]e^(kt))}}}, with:
{{{A}}}  being remaining amount after time t (115, in this case)
{{{A[o]}}} being Original/Initial amount (6,000, in this case)
{{{k}}}  being the constant (k > 0 signifies RATE OF GROWTH ;  k < 0 signifies RATE OF DECAY ; k = - .0385, in this case) 
{{{t}}}  being time, in stated periods (Unknown, in this case)

                                          {{{matrix(1,3, A, "=", A[o]e^(kt))}}}
                                        {{{matrix(1,3, 115, "=", "6,000"e^(- .0385t))}}} ----- Substituting 115 for A, 6,000 for {{{A[o]}}}, and - .0385 for k
                                      {{{matrix(2,3, 115/"6,000", "=", e^(- .0385t), 23/"1,200", "=", e^(- .0385t))}}}
                                   {{{matrix(1,3, - .0385t, "=", ln (23/"1,200"))}}} ------ Converting to LOGARITHMIC (Natural) form
Time it takes for 115 grams to remain, or {{{highlight_green(matrix(1,8, t, "=", ln (23/"1,200")/(- .0385), "=", 102.716, or, highlight(102.72), years))}}}

The correct answer should actually be in WHOLE-NUMBER years (103 to be specific)!</pre>