Question 1171695
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The phrasing "is reduced by 50%" is the same as "reduced by half". Every 5 hours, the amount of caffeine is reduced by half.


If you started with 320 mg of caffeine, then you'll have 320/2 = 160 mg of caffeine after 5 hours. Then you'll have 160/2 = 80 mg after another 5 hours (5+5 = 10 total so far)


We keep dividing the amount in half and keep track how much time has passed by (along with the number of half-lives)


A table is useful for this sort of thing.
<table border = "1" cellpadding = "5"><tr><td>Half-life </td><td>Time</td><td>Amount</td></tr><tr><td>0</td><td>0</td><td>320</td></tr><tr><td>1</td><td>5</td><td>160</td></tr><tr><td>2</td><td>10</td><td>80</td></tr><tr><td>3</td><td>15</td><td>40</td></tr><tr><td>4</td><td>20</td><td>20</td></tr><tr><td>5</td><td>25</td><td>10</td></tr><tr><td>6</td><td>30</td><td>5</td></tr></table>
The time value is in hours, and the amount represents the amount of caffeine in mg.


The table shows at time 30 hours, there's exactly 5 mg of caffeine left in the bloodstream.
This happens after 6 half-lives have occurred.


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An algebraic approach


x = number of half-lives that pass by 
y = amount of caffeine left over in the bloodstream


The equation tying together x and y is
y = 320*(0.5)^x
where 1/2 = 0.5 represents the multiplier indicating a 50% reduction


We could guess-and-check by plugging in random values of x to see if we get y = 5.
But let's solve for x using algebra


Plug in y = 5 and use logs to isolate the exponent
y = 320*(0.5)^x
5 = 320*(0.5)^x
5/320 = (0.5)^x
0.015625 = (0.5)^x
log(0.015625) = log( (0.5)^x )
log(0.015625) = x*log( 0.5 )
x*log( 0.5 ) = log(0.015625)
x = log(0.015625)/log(0.5)
x = 6
This helps confirm the half-life we got earlier with the table.


Another confirmation method is to plug x = 6 into the equation
y = 320*(0.5)^x
y = 320*(0.5)^6
y = 320*0.015625
y = 5


x = 6 half-lives occurring mean 5x = 5*6 = 30 hours have passed by (since each half-life is 5 hours).


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Answer: 30 hours
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