$$ \leftarrow $$ Use the exponential decay model, $$ \mathrm{A}=\mathrm{A}_{0} e^{\mathrm{kt}} $$, to solve the following. The half-life of a certain substance is 16 years. How long will it take for a sample of this substance to decay to $$ 75 \% $$ of its original amount? It will take approximately original amount. (Round the final answer to on places as needed.) years percent

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**Step 1. Determine the decay constant $$ k $$ using the half-life**

For a half-life of 16 years, the amount decays to $$ \frac{1}{2} $$ of its original value in 16 years:
$$
\frac{A_0}{2} = A_0 e^{16k}
$$
Divide both sides by $$ A_0 $$:
$$
\frac{1}{2} = e^{16k}
$$
Take the natural logarithm on both sides:
$$
\ln\left(\frac{1}{2}\right) = 16k
$$
Solve for $$ k $$:
$$
k = \frac{\ln\left(\frac{1}{2}\right)}{16} = -\frac{\ln 2}{16}
$$

**Step 2. Use the decay model to determine the time for decay to $$ 75\% $$ of the original amount**

We have:
$$
A = 0.75\,A_{0} = A_{0} e^{kt}
$$
Divide both sides by $$ A_{0} $$:
$$
0.75 = e^{kt}
$$
Take the natural logarithm of both sides:
$$
\ln(0.75) = kt
$$
Substitute the value of $$ k $$:
$$
\ln(0.75) = \left(-\frac{\ln 2}{16}\right) t
$$
Solve for $$ t $$:
$$
t = \frac{\ln(0.75)}{-\frac{\ln 2}{16}} = -16\,\frac{\ln(0.75)}{\ln 2}
$$

**Step 3. Compute the numerical value**

Approximate the logarithms:
$$
\ln(0.75) \approx -0.28768, \quad \ln(2) \approx 0.69315
$$
Thus:
$$
t \approx -16\,\frac{-0.28768}{0.69315} \approx 16 \times 0.4159 \approx 6.65
$$

**Final Answer**

It will take approximately $$ 6.65 $$ years for the substance to decay to $$ 75\% $$ of its original amount.
It will take approximately 6.65 years for the substance to decay to 75% of its original amount.

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