Question

a baseball card sold for $205 in 1979 and was sold again in 1985 for $488. assume that the growth in the value v of the collectors item was exponential.
a) find the value k of the exponential growth rate. assume ( v_0 = 205 ).
k = 0.0488
(round to the nearest thousandth.)

Explanation:

Step1: Write the exponential growth formula

The exponential growth formula is \(V = V_0e^{kt}\). Here, \(V_0 = 205\), \(V = 488\), and \(t=1985 - 1979=6\) years.
Substituting the values into the formula gives \(488 = 205e^{6k}\).

Step2: Solve for \(e^{6k}\)

Divide both sides of the equation \(488 = 205e^{6k}\) by \(205\):
\(\frac{488}{205}=e^{6k}\).
Since \(\frac{488}{205} = 2.3804878\), the equation becomes \(2.3804878=e^{6k}\).

Step3: Take the natural logarithm of both sides

Take the natural logarithm of both sides: \(\ln(2.3804878)=\ln(e^{6k})\).
Using the property \(\ln(e^{x})=x\), we get \(\ln(2.3804878) = 6k\).
Since \(\ln(2.3804878)\approx0.867\), then \(0.867 = 6k\).

Step4: Solve for \(k\)

Divide both sides of \(0.867 = 6k\) by \(6\): \(k=\frac{0.867}{6}\).
\(k = 0.1445\approx0.145\)

Answer:

\(k\approx0.145\)