Question

Use the formula R = 6e12.77x, where x is the blood alcohol concentration and R, as a percent, is the risk of having a car accident. What blood alcohol concentration corresponds to a 50% risk of a car accident?

Answer 1

$R=6e^{12.77x}$
$50=6e^{12.77x}$
$\frac{50}{6}= e^{12.77x}$
$log \frac{50}{6}=log( e^{12.77x})$
$0.920818754=12.77x$
$x=0.07$

Answer 2

Answer:

0.1660 ( approx )

Step-by-step explanation:

Given,

The percentage of risk of having a car accident is,

$R=6e^{12.77x}----(1)$

Where, x is the blood alcohol concentration,

From equation (1),

$\frac{R}{6}=e^{12.77x}$

Taking natural log both sides,

$ln(\frac{R}{6})=ln(e^{12.77x})$

$ln(\frac{R}{6})=12.77xln(e)$

Since, ln(e) = 1,

$ln(\frac{R}{6})=12.77x$

$\implies x = \frac{ln(\frac{R}{6})}{12.77}$

R = 50 %,

$\implies x=\frac{ln(\frac{50}{6})}{12.77}$

$=\frac{2.1202635362}{12.77}$

$=0.16603473267\approx 0.1660$