A valve on a storage tank is opened for 4 hours to release a chemical in a manufacturing process. The flow rate R (in liters per hour) at time t (in hours) is given by the linear Model $\ln R = - 0.6155 t + 15.0609$ Write the linear model in exponential form.
A) $R = - 15.0609 e ^ { - 0.6155 t }$
B) $R = 8.0609 e ^ { - 0.6155 t }$
C) $R = 15.0609 e ^ { - 0.6155 t }$
D) $R = 3,474,287.5695 e ^ { - 0.6155 t }$
E) $R = 3,474,287.5695 e ^ { 0.6155 t }$

Question

A valve on a storage tank is opened for 4 hours to release a chemical in a manufacturing process. The flow rate R (in liters per hour) at time t (in hours) is given by the linear Model $\ln R = - 0.6155 t + 15.0609$ Write the linear model in exponential form. 
A) $R = - 15.0609 e ^ { - 0.6155 t }$
B) $R = 8.0609 e ^ { - 0.6155 t }$
C) $R = 15.0609 e ^ { - 0.6155 t }$
D) $R = 3,474,287.5695 e ^ { - 0.6155 t }$
E) $R = 3,474,287.5695 e ^ { 0.6155 t }$

Quizplus · Accepted Answer

To convert the given logarithmic model $\ln R = - 0.6155 t + 15.0609$ to exponential form, we use the property that if $\ln x = y$, then $x = e^y$. Applying this to the given model, we get $R = e^{-0.6155t + 15.0609}$. Simplifying further, $R = e^{15.0609} \cdot e^{-0.6155t}$. Calculating $e^{15.0609}$ gives approximately $3,474,287.5695$, so the correct model in exponential form is $R = 3,474,287.5695 e^{-0.6155t}$, which matches option D.

A Valve on a Storage Tank Is Opened for 4 ln⁡R=−0.6155t+15.0609\ln R = - 0.6155 t + 15.0609lnR=−0.6155t+15.0609

A Valve on a Storage Tank Is Opened for 4 $\ln R = - 0.6155 t + 15.0609$