To find the Celsius temperature at which 15 grams of the chemical will dissolve, we use the given exponential model [tex]\( A(t) = 10 e^{0.0095 t} \)[/tex]. We need to determine the temperature [tex]\( t \)[/tex] at which [tex]\( A(t) = 15 \)[/tex].
Here are the steps to find [tex]\( t \)[/tex]:
1. Set up the equation:
We start with the model:
[tex]\[
A(t) = 10 e^{0.0095 t}
\][/tex]
We need to find [tex]\( t \)[/tex] for which [tex]\( A(t) = 15 \)[/tex]:
[tex]\[
15 = 10 e^{0.0095 t}
\][/tex]
2. Isolate the exponential term:
Divide both sides of the equation by 10:
[tex]\[
\frac{15}{10} = e^{0.0095 t}
\][/tex]
Simplify:
[tex]\[
1.5 = e^{0.0095 t}
\][/tex]
3. Solve for [tex]\( t \)[/tex]:
To get [tex]\( t \)[/tex] out of the exponent, take the natural logarithm of both sides:
[tex]\[
\ln(1.5) = \ln(e^{0.0095 t})
\][/tex]
Using the property of logarithms [tex]\( \ln(e^x) = x \)[/tex], we have:
[tex]\[
\ln(1.5) = 0.0095 t
\][/tex]
4. Isolate [tex]\( t \)[/tex]:
Divide both sides by 0.0095:
[tex]\[
t = \frac{\ln(1.5)}{0.0095}
\][/tex]
5. Calculate [tex]\( t \)[/tex]:
Perform the calculation:
[tex]\[
t \approx 42.68053769559625
\][/tex]
6. Round to the nearest degree:
To report the temperature to the nearest degree:
[tex]\[
t \approx 43
\][/tex]
So, the temperature at which 15 grams of the chemical will dissolve is approximately [tex]\( 43 \)[/tex] degrees Celsius.
