Question

question 16 (mandatory) (1 point) a student records the temperature of an art project that has been left to cool on a counter. the room temperature is 18°c. an equation that models this situation is t(t)=59(0.5)^(t/10)+18 where t is the temperature in degrees celsius and t is the time in minutes. determine the temperature, to the nearest degree, of the art project after 10 minutes. how much time did it take for the project to reach a temperature of 30°c?

Explanation:

Step1: Set up the equation

We are given the temperature - time equation $T(t)=59(0.5)^{\frac{t}{10}}+18$, and we want to find $t$ when $T(t) = 30$. So we set up the equation $30=59(0.5)^{\frac{t}{10}}+18$.

Step2: Isolate the exponential term

First, subtract 18 from both sides of the equation:
$30 - 18=59(0.5)^{\frac{t}{10}}$, which simplifies to $12 = 59(0.5)^{\frac{t}{10}}$. Then, divide both sides by 59: $\frac{12}{59}=(0.5)^{\frac{t}{10}}$.

Step3: Take the natural - logarithm of both sides

$\ln(\frac{12}{59})=\ln((0.5)^{\frac{t}{10}})$. Using the property of logarithms $\ln(a^b)=b\ln(a)$, we get $\ln(\frac{12}{59})=\frac{t}{10}\ln(0.5)$.

Step4: Solve for $t$

First, find $\ln(\frac{12}{59})\approx\ln(12)-\ln(59)\approx2.4849 - 4.0775=-1.5926$ and $\ln(0.5)\approx - 0.6931$. Then, $t = 10\times\frac{\ln(\frac{12}{59})}{\ln(0.5)}$. Substitute the values of the logarithms: $t = 10\times\frac{-1.5926}{-0.6931}\approx23$.

Answer:

It takes approximately 23 minutes for the art - project to reach a temperature of $30^{\circ}C$.