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The population of MeadowView is increasing at a rate of 3% per year. If the population is 100,000 today, what will it be in three years?
This is exponential growth: the population increases by a certain percentage every year.
One year later the population becomes \(100000 \times (1+0.03) = 100000 \times 1.03 \).
Two years later it becomes \(100000 \times 1.03 \times (1+0.03) = 100000 \times 1.03^2 \)
Three years later it becomes \(100000 \times 1.03^2 \times (1+0.03) = 100000 \times 1.03^3 = 109273 \)
The pattern \(n\) years later the population is \(100000 \times 1.03^n \)
is pretty clear.
For example, 30 years later, the population becomes
\(100000 \times 1.03^{30} = 242726 \)
The price of a gallon of gasoline is increasing by 4 cents per week. If the price is \( \$ 3.10 \) per gallon today, what will it be in ten weeks?
This is linear growth: the price increases by a certain amount (4 cents) every week independently on what it was.
During ten weeks the price will increase by
\(4 \times 10=40 \),
and will become \(3.10 + 40 = \$ 3.50 \)
Here the pattern is: $n$ weeks later the price is
\(310 + 4 \times n\) (cents).
The value of your car is decreasing by 10% per year. If the car is worth \(\$ 12,000\) today, what will it worth in two years?
This is exponential decay: the value decreases by a certain percentage (10%) every year.
One year from now the car is worth
\(12000 - 12000 \times 0.1 = 12000 \times (1-0.1) = 12000 \times 0.9 = \$ 10800 \)
Two years from now the car is worth
\(10800 - 10800 \times 0.1 = 10800-1080 = \$9720 \)
One year from now the car is worth
\(12000 - 12000 \times 0.1 = 12000 \times (1-0.1) = 12000 \times 0.9 = \$ 10800 \)
Two years from now the car is worth
\( (12000 \times (1-0.1))\times(1-0.1) = 12000 \times (1-0.1)^2 \)
Three years from now the car is worth
\( (12000 \times (1-0.1)^2)\times(1-0.1) = 12000 \times (1-0.1)^3 \)
Now the pattern becomes clear: $n$ years from now the car is worth
\( 12000 \times (1-0.1)^n = 12000 \times 0.9^n \)
For example, $20$ years from now, the car is worth
\( 12000 \times 0.9^{20} \approx \$ 1,459\)
How many grains of wheat should be placed on square 32 of the chessboard?
Solution. Recall that there is 1 grain on the first square.
there are \(2=2^1\) on the second square, \(4=2^2\) on the third square, \(8=2^3\) on the fourth, and so on...
The pattern is clear: square number \(n\) has \(2^{n-1}\) grains.
In particular, square 32 has
\(2^{31} =2147483648\approx 2.1 \times 10^9 \) grains.
Find the total number of grains and their weight at this point assuming that a grain of wheat weights 1/7000 pound.
The pattern for the total number of grains on board is only a bit more complicated, and is given on p475 of the textbook.
After square \(n\) is full, there are \(2^n-1\) grains on the board.
For \( n=32\), this is \(2^{32}-1 = 4,294,967,295 \approx 4.2 \times 10^9\)
The weight in question is
\(4294967295 \times 1/7000 \approx 613,567 \) pounds.
Suppose that you could keep making a single stack of the pennies. After how many days would the stack be long enough to reach the nearest star (beyond the Sun), which is \( 4 \times 10^{13} \) km away?
There are \( 2^t \) pennies after \(t\) days.
We need the thickness of a penny let's try to find it out
We thus want that
\(2^t \approx \frac{4 \times 10^{16}}{1.52 \times 10^{-3}} \approx 2.6 \times 10^{19} \)
After some trial and error we find that
\(2^{65} \approx 3.7 \times 10^{19} \)
so 65 days should suffice.
How many bacteria are there in the bottle at 11:15?
Solution. Recall that \(t\) minutes past 11:00, there are \(2^t\) bacteria in the bottle.
Thus there are \(2^{15}\) of them at 11:15.
What fraction of the bottle is full at that time?
The bottle is full at noon with \(2^{60}\) bacteria.
At 11:15, \(2^{15}\) bacteria constitute
\( \frac{2^{15}}{2^{60}} = \frac{1}{2^{45}} = 2^{-45} \approx 2.8 \times 10^{-14} \)