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After one year in the bank, Raphael's deposit will be worth $1.05 \times 10,000$ dollars (or \$10,050). After two years we will multiply by 1.05 again because in the second year interest is earned not only on the initial \$10,000 deposit but also on the \$50 interest earned in the first year: so after two years the account will have a balance of $1.05\times 1.05 \times 10,000$ dollars. Each successive year, we will multiply by 1.05 again, reflecting a 5% earnings for the amount of money in the account at the beginning of the year. So after $t$ years, Raphael will have $1.05^t \times 10.000$ dollars in the account. To find out when he will have \$50,000 in the account we solve the equation $$1.05^t \times 10,000 = 50,000.$$ Dividing both sides by 10,000 and taking the natural log gives $t = \frac{\ln{5}}{\ln{1.05}} \approx 33$. It will take about 33 years for Raphael's bank account balance to reach \$50,000.
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In this case, we are given that for each meter traveled on the road up the Alpe d'Huez, the elevation goes up by about 0.08 meters. If we let $x$ denote the number of meters traveled on the road and $f(x)$ the elevation at that spot on the road then we have $$f(x) = 740 + 0.08x:$$ when $x = 0$ we are at the bottom of the climb and each increment of 1 in $x$ results in an elevation gain of 0.08 meters. Ten kilometers is 10,000 meters so the elevation ten kilometers into the climb is given by $f(10,000) = 740 + 800 = 1540$. The elevation 10 kilometers into the climb is about 1540 meters. The total distance of the climb is 14 kilometers or 14,000 meters so the elevation at the top of the climb is $$f(14,000) = 740 +1120 = 1860,$$ just a little under 2000 meters.
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Suppose $g(t)$ represents the amount of Radon 222 remaining, measured in grams, from our sample after $t$ days. Since Radon decays exponentially we will use an exponential function with negative growth rate for our model: $g(t) = ae^{-ct}$ where $c \gt 0$ gives the rate of decay. Since $e^0 = 1$ we have $g(0) = a$. We know that 100 grams are in the initial sample of Radon so $g(0) = 100$ and this means that $a = 100$. So $g(t) = 100e^{-ct}$. We are also given that after 3.8 days only 50 grams of the Radon remain so $g(3.8) = 50$. Substituting 3.8 into $g(t)$ gives $$50 = 100e^{-3.8c}$$ or $c = \frac{\ln{\frac{1}{2}}}{-3.8}$ so $c$ is about 0.18 giving us a model of $g(t) = 100e^{-0.18t}$ for the amount of Radon left after $t$ days. We want to know when 1 gram is left so we need to solve $100e^{-0.18t} = 1$. This gives $t = \frac{\ln{\left(\frac{1}{100}\right)}}{-0.18} \approx 26$. It will take about 26 days before there is only 1 gram of Radon 222 remaining from the 100 gram sample.
