A certain element decays at a constant rate of \( 3 \% \) per year, If you start with 15 grams of the element, how long will it take before there are only three grams left? a.) 52.8 years b.) 51.8 years c.) 4.51 years d.) 54.4 years

To solve this problem, we can use the formula for exponential decay: \[ A = P \times (1 - r)^t \] where: - \( A \) is the amount of the element left after time \( t \), - \( P \) is the initial amount of the element (15 grams in this case), - \( r \) is the decay rate (3% or 0.03 in decimal form), - \( t \) is the time in years. We are given that the initial amount of the element is 15 grams and we want to find out how long it will take for there to be only 3 grams left. We can set up the equation as follows: \[ 3 = 15 \times (1 - 0.03)^t \] Now, we can solve for \( t \) using the formula for exponential decay. Solve the equation by following steps: - step0: Solve for \(t\): \(3=15\left(1-0.03\right)^{t}\) - step1: Simplify: \(3=15\left(\frac{97}{100}\right)^{t}\) - step2: Swap the sides: \(15\left(\frac{97}{100}\right)^{t}=3\) - step3: Multiply by the reciprocal: \(15\left(\frac{97}{100}\right)^{t}\times \frac{1}{15}=3\times \frac{1}{15}\) - step4: Multiply: \(\left(\frac{97}{100}\right)^{t}=\frac{1}{5}\) - step5: Take the logarithm of both sides: \(\log_{\frac{97}{100}}{\left(\left(\frac{97}{100}\right)^{t}\right)}=\log_{\frac{97}{100}}{\left(\frac{1}{5}\right)}\) - step6: Evaluate the logarithm: \(t=\log_{\frac{97}{100}}{\left(\frac{1}{5}\right)}\) - step7: Simplify: \(t=-\log_{\frac{97}{100}}{\left(5\right)}\) The solution to the equation is \( t = -\log_{\frac{97}{100}}{(5)} \). Now, we can calculate the value of \( t \) to determine how long it will take for there to be only 3 grams left. Calculate the value by following steps: - step0: Calculate: \(-\log_{\frac{97}{100}}{\left(5\right)}\) The time it will take for there to be only 3 grams left is approximately 52.84 years. Therefore, the correct answer is a.) 52.8 years.