Question

1. a baker records the internal temperature of a pie that has been left to cool on a counter. the room temperature is 14°c. an equation that models this situation is (t(t)=68(0.5)^{\frac{t}{10}} + 14) where (t) is the temperature in degrees celsius and (t) is the time in minutes. a) determine the temperature, to the nearest degree, of the pie after 30 minutes. b) how much time did it take for the pie to reach an internal temperature of 34°c?

Explanation:

Step1: Substitute $t = 30$ into the formula

$T(30)=68(0.5)^{\frac{30}{10}}+14$

Step2: Simplify the exponent

$(0.5)^{\frac{30}{10}}=(0.5)^3$

Step3: Calculate $(0.5)^3$

$(0.5)^3 = 0.5\times0.5\times0.5=0.125$

Step4: Calculate $68\times(0.5)^3$

$68\times0.125 = 8.5$

Step5: Calculate $T(30)$

$T(30)=8.5 + 14=22.5\approx23$

For part b:

Step1: Set $T(t)=34$ in the formula

$34=68(0.5)^{\frac{t}{10}}+14$

Step2: Subtract 14 from both sides

$34 - 14=68(0.5)^{\frac{t}{10}}$
$20 = 68(0.5)^{\frac{t}{10}}$

Step3: Divide both sides by 68

$(0.5)^{\frac{t}{10}}=\frac{20}{68}=\frac{5}{17}$

Step4: Take the logarithm of both sides

$\log(0.5)^{\frac{t}{10}}=\log(\frac{5}{17})$
$\frac{t}{10}\log(0.5)=\log(5)-\log(17)$

Step5: Solve for $t$

$t = 10\times\frac{\log(5)-\log(17)}{\log(0.5)}$
$t\approx10\times\frac{0.69897 - 1.23045}{- 0.30103}$
$t\approx10\times\frac{- 0.53148}{- 0.30103}\approx17.65\approx18$

Answer:

a) $23^{\circ}C$
b) $18$ minutes