Question 839175
Look at this problem, how you say, taking 12% repeatedly; but LOOK FOR THE PATTERN.  You can symbolize that pattern and put it into an equation.


Time________________Caffeine Still Present
0___________________120 mg.
1___________________120-120(0.12)
2___________________(120-120(0.12))-......
I see what you mean.



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Try from another point of view.  The amount REMAINING after each hour is 100-12 percent; and this percentage is  applied each hour.
100-12=88.
Instead of examining how 12% of the previous caffeine is eliminated each hour, look at the 88% of caffeine which still remains after every hour.


Time__________Ceffeine Still Present
0_____________120
1_____________{{{120*0.88}}}
2_____________{{{120*0.88*0.88}}}
3_____________{{{120*0.88*0.88*0.88}}}
4_____________{{{120*0.88*0.88*0.88*0.88}}}
5_____________{{{120(0.88)^5}}}, you should see the pattern already, and using exponents make the notation more compact.


{{{highlight(C(t)=120(0.88)^t)}}}, function notation chosen.  Amount of caffeine remaining after {{{t}}} number of hours, is {{{C(t)}}}.  This is exponential decay, the rate being loss of 12% each hour.  Initial amount at time zero is 120 milligrams.


Your question then becomes, find {{{t}}} when {{{C(t)=10}}}.
You start with this:  {{{highlight(10=120(0.88)^t)}}} and you want to solve for {{{t}}}.



----------------SOLVING FOR t------------
{{{10/120=0.88^t}}}
{{{1/12=0.88t}}}
{{{log(10,1/12)=log(10,0.88^t)}}}
{{{log(10,1/12)=t*log(10,0.88)}}}
{{{t=log(1/12)/log(0.88)}}}
{{{t=(-1.07918)/(-0.05552)}}}
{{{highlight(t=19.4)}}} hours = 19 hours 30 minutes