A painting sold for $\$ 210$ in 1979 and was sold again in 1989 for $\$ 500$. Assume that the growth in the value $V$ of the collector's item was exponential. a) Find the value $\mathrm{k}$ of the exponential growth rate. Assume $\mathrm{V}_{\mathrm{o}}=210$. \[ \mathrm{k}=\square \] (Round to the nearest thousandth.)

Step 1 ：The value of the painting is growing exponentially, which means it follows the formula \(V = V_0 \cdot e^{kt}\), where \(V\) is the final value, \(V_0\) is the initial value, \(k\) is the rate of growth, \(t\) is the time, and \(e\) is the base of the natural logarithm.

Step 2 ：We know \(V\), \(V_0\), and \(t\), so we can solve for \(k\). The time \(t\) is the difference in years between 1989 and 1979, which is 10 years.

Step 3 ：Substitute these values into the formula: \(V = 500\), \(V_0 = 210\), \(t = 10\).

Step 4 ：Solving for \(k\) gives \(k = 0.0867500567704723\).

Step 5 ：Rounding to the nearest thousandth, the value of the exponential growth rate \(k\) is approximately \(\boxed{0.087}\).