Question 320172
Please help me solve this word problem: A certain radioactive element decays at a rate of 4% per year. Find the half-life of the substance (i.e. the time it will take for one half of any given amount of the substance to decay). 
<pre><font size = 3 color = "indigo"><b>
Let {{{A}}} be the original amount

After 1 year only 96% will remain, which will be {{{0.96A}}}
After 2 years only 96% of {{{0.96A}}} will remain, which is {{{0.96^2*A}}}
After 3 years only 96% of {{{0.96^2*A}}} will remain, which is {{{0.96^3*A}}}
...
After n years only 96% of {{{0.96^(n-1)A}}} will remain, which is {{{0.96^n*A}}}

We want to know how many years before what will remain will be only {{{1/2}}}{{{A}}} or {{{0.5A}}} 

{{{0.96^n*A}}}{{{"="}}}{{{0.5A}}}

Divide both sides by A

{{{0.96^n}}}{{{"="}}}{{{0.5}}}

Take logs of both sides:

{{{log((0.96^n))}}}{{{"="}}}{{{log((0.5))}}}

Use the rule of logs that says {{{log(B,(A^C))=C*log(B,(A))}}} to bring
the exponent n in front of the log

{{{n*log((0.96))}}}{{{"="}}}{{{log((0.5))}}}

Divide both sides by {{{log((0.96))}}}

{{{n}}}{{{"="}}}{{{log((0.5))/log((0.96))}}}{{{"="}}}{{{16.97974802}}}

or 17 years, rounded to the first whole year that the amount will be
no more than half the original amount.

Edwin</pre>