Step 1 :Setting \(P(t)\) to \(0.77P_0\) (since the bone has lost 23% of its carbon-14, it retains 77% or 0.77 of its original amount), we find the following: \[0.77P_0 = P_0 e^{-0.00012097t}\]
Step 2 :Divide both sides by \(P_0\) to get: \[0.77 = e^{-0.00012097t}\]
Step 3 :Take the natural logarithm of both sides to isolate the exponent: \[\ln(0.77) = -0.00012097t\]
Step 4 :Divide both sides by \(-0.00012097\) to solve for \(t\): \[t = \frac{\ln(0.77)}{-0.00012097}\]
Step 5 :Calculate the value of \(t\) to find the age of the bone: \[t \approx 3030\] years
Step 6 :Our final answer is \(\boxed{3030}\) years.