Question

The concentration of alcohol in a person's bloodstream is measurable. Suppose that the relative risk \(R\) of having an accident while driving a car can be modeled by an equation of the form$$R=e^{k x}$$ where \(x\) is the percent concentration of alcohol in the bloodstream and \(k\) is a constant. (a) Suppose that a concentration of alcohol in the bloodstream of 0.03 percent results in a relative risk of an accident of \(1.4 .\) Find the constant \(k\) in the equation. (b) Using this value of \(k,\) what is the relative risk if the concentration is 0.17 percent? (c) Using the same value of \(k,\) what concentration of alcohol corresponds to a relative risk of \(100 ?\) (d) If the law asserts that anyone with a relative risk of having an accident of 5 or more should not have driving privileges, at what concentration of alcohol in the bloodstream should a driver be arrested and charged with a DUI? (e) Compare this situation with that of Example \(10 .\) If you were a lawmaker, which situation would you support? Give your reasons.

Short answer

k ≈ 11.2157. Relative risk at 0.17%: ≈ 6.725. Concentration for risk of 100: ≈ 0.41%. Concentration for arrest with risk of 5: ≈ 0.14%.

Step-by-step solution

- Find the constant k

Given the equation for relative risk: $$R = e^{kx}$$, use the provided information that a concentration of 0.03 percent results in a relative risk of 1.4. Substitute the given values into the equation: $$1.4 = e^{k \times 0.03}.$$ Next, solve for k by taking the natural logarithm of both sides: $$\text{ln}(1.4) = k \times 0.03.$$ Finally, solve for k:$$k = \frac{\text{ln}(1.4)}{0.03}.$$

- Calculate k

Using the natural logarithm, find: $$\text{ln}(1.4) \thickapprox 0.33647.$$ Then: $$k = \frac{0.33647}{0.03} \thickapprox 11.2157.$$

- Find the relative risk for 0.17 percent concentration

Using the above value of k, substitute into the equation: $$R = e^{kx}.$$ For a concentration of 0.17 percent, substitute x = 0.17: $$R = e^{11.2157 \times 0.17}.$$

- Calculate the relative risk for 0.17 percent concentration

Calculate the power first: $$11.2157 \times 0.17 \thickapprox 1.90667.$$ Then find: $$R = e^{1.90667} \thickapprox 6.725.$$

- Find the concentration for a relative risk of 100

For a given relative risk R = 100, substitute into the equation: $$100 = e^{11.2157x}.$$ Take the natural logarithm of both sides: $$\text{ln}(100) = 11.2157x.$$ Solve for x: $$x = \frac{\text{ln}(100)}{11.2157}.$$

- Calculate the concentration for a relative risk of 100

Using the natural logarithm: $$\text{ln}(100) \thickapprox 4.60517.$$ Then: $$x = \frac{4.60517}{11.2157} \thickapprox 0.4107.$$ So, a concentration of approximately 0.41 percent results in a relative risk of 100.

- Find the concentration corresponding to a relative risk of 5

Given a relative risk R = 5, substitute into the equation: $$5 = e^{11.2157x}.$$ Taking the natural logarithm of both sides: $$\text{ln}(5) = 11.2157x.$$ Solve for x: $$x = \frac{\text{ln}(5)}{11.2157}.$$

- Calculate the concentration for a relative risk of 5

Using the natural logarithm: $$\text{ln}(5) \thickapprox 1.60944.$$ Then: $$x = \frac{1.60944}{11.2157} \thickapprox 0.1436.$$ A concentration of approximately 0.14 percent results in a relative risk of 5.

Conclusion: Comparison to Example 10

In Example 10, a specific scenario is likely provided. As a lawmaker, one might support or reject the situation based on additional data or social factors. It's important to balance road safety and fair legal thresholds. Always provide an analytical approach covering the impact on societal health and accident prevention.

Key concepts

Exponential Equations

Exponential equations involve variables in their exponents. They often model real-life scenarios like population growth or decay, radiation, and in this case, blood alcohol concentration influencing accident risk. These equations take the form: $$R = e^{kx}$$ Here, the base of the natural logarithms, e, is a constant approximately equal to 2.71828. The variable k and x can change depending on the situation.
To solve exponential equations, it is often necessary to use the natural logarithm. This is because taking the natural log of an exponential term simplifies it significantly. For example:
$$ \text{If } R = e^{kx}, \text{ then taking the natural log (ln) of both sides gives: } \text{ln}(R) = kx.$$
Studying the properties of exponential functions helps us predict future trends when the current and initial states are known. In this exercise, we use the exponential relationship to find the constant k and predict risks.

Natural Logarithms

Natural logarithms simplify solving exponential equations. The natural logarithm, denoted as ln, has a base of e (approximately 2.71828). It helps us solve for variables in exponent positions.
For instance, if we have the equation: $$e^{kx} = R,$$ applying the natural logarithm to both sides allows us to isolate kx: $$\text{ln}(e^{kx}) = \text{ln}(R).$$
Due to logarithmic properties, $$\text{ln}(e^{kx}) = kx,$$ which simplifies our equation to: $$kx = \text{ln}(R).$$
Using this, we can solve for k or x. For example: Suppose we know: As given: $$1.4 = e^{k \times 0.03},$$ then: $$ \text{ln}(1.4) = k \times 0.03.$$ We solve for k by rearranging: $$ k = \frac{\text{ln}(1.4)}{0.03}.$$ This fundamental technique unlocks exponential equations by transforming them into linear form using logarithms.

Alcohol Concentration

Blood alcohol concentration (BAC) measures the amount of alcohol in a person's bloodstream. This concentration can significantly affect driving abilities, and it's often used to enforce legal limits on drinking and driving.
BAC is generally expressed as a percentage. For instance, a BAC of 0.03% means there are 0.03 grams of alcohol per 100 milliliters of blood. Legal limits are usually below 0.08%, but even lower concentrations can impair driving abilities.
In the context of risk assessment, higher BAC levels correlate with higher relative risks (R) of accidents. That's why we use equations like: $$R = e^{kx}$$ to model this relationship.
For example, with a BAC of 0.17%, the relative risk skyrockets to approximately 6.725 times the risk of a sober driver. Understanding this relationship aids in setting reasonable legal limits to enhance road safety. By calculating such risks, we determine safer guidelines and legal thresholds, like setting a limit where the relative risk reaches a dangerous level, leading to DUI charges to prevent potential accidents.