To determine how many minutes Greg should leave the bottle in the cooler, we need to find the time [tex]\( t \)[/tex] when the temperature [tex]\( C(t) \)[/tex] reaches 15 degrees Celsius. The temperature function is given by:
[tex]\[ C(t) = 5 + 13e^{-0.034t} \][/tex]
We need to solve for [tex]\( t \)[/tex] when [tex]\( C(t) = 15 \)[/tex]. Let's set up the equation:
[tex]\[ 15 = 5 + 13e^{-0.034t} \][/tex]
Subtract 5 from both sides to isolate the exponential term:
[tex]\[ 15 - 5 = 13e^{-0.034t} \][/tex]
This simplifies to:
[tex]\[ 10 = 13e^{-0.034t} \][/tex]
Next, divide both sides by 13 to further isolate the exponential term:
[tex]\[ \frac{10}{13} = e^{-0.034t} \][/tex]
To solve for [tex]\( t \)[/tex], we need to take the natural logarithm (ln) of both sides. The natural logarithm will allow us to isolate the exponent. Applying the natural logarithm, we get:
[tex]\[ \ln\left(\frac{10}{13}\right) = \ln\left(e^{-0.034t}\right) \][/tex]
Using the property of logarithms that [tex]\( \ln(e^x) = x \)[/tex], the equation simplifies to:
[tex]\[ \ln\left(\frac{10}{13}\right) = -0.034t \][/tex]
To solve for [tex]\( t \)[/tex], divide both sides of the equation by [tex]\(-0.034\)[/tex]:
[tex]\[ t = \frac{\ln\left(10/13\right)}{-0.034} \][/tex]
Now, let's find the value of [tex]\( t \)[/tex] by calculating the natural logarithm and division precisely. The natural logarithm of [tex]\( 10/13 \)[/tex] is approximately -0.262364264467.
Then,
[tex]\[ t \approx \frac{-0.262364264467}{-0.034} \][/tex]
Performing the division, we get:
[tex]\[ t \approx 7.716596013749735 \][/tex]
Rounding this result to the nearest tenth, we obtain:
[tex]\[ t \approx 7.7 \][/tex]
Therefore, Greg should leave the bottle in the cooler for approximately [tex]\( 7.7 \)[/tex] minutes to reach a temperature of 15 degrees Celsius.
