Questions: The values of y, y0, and t are given. For an equation of the form y=y0 e^(kt), give the exact value of k in terms of natural logarithms. y0=120, t=3, y=40 k=.

Solution Steps

To find the value of \( k \) in the equation \( y = y_0 e^{kt} \), we can rearrange the equation to solve for \( k \). Start by dividing both sides by \( y_0 \) to isolate the exponential term. Then, take the natural logarithm of both sides to solve for \( k \).

We start with the equation given in the problem: \[ y = y_0 e^{kt} \] To isolate \( k \), we first divide both sides by \( y_0 \): \[ \frac{y}{y_0} = e^{kt} \]

Next, we take the natural logarithm of both sides: \[ \ln\left(\frac{y}{y_0}\right) = kt \]

Now, we can solve for \( k \) by dividing both sides by \( t \): \[ k = \frac{\ln\left(\frac{y}{y_0}\right)}{t} \]

Substituting the given values \( y_0 = 120 \), \( t = 3 \), and \( y = 40 \): \[ k = \frac{\ln\left(\frac{40}{120}\right)}{3} = \frac{\ln\left(\frac{1}{3}\right)}{3} \]

Evaluating the expression gives: \[ k \approx -0.3662 \]

Final Answer