The amount of​ carbon-14 present in animal bones after t years is given by ​P(t)equalsUpper P 0 e Superscript negative 0.00012 t. A bone has lost 19​% of its​ carbon-14. How old is the​ bone?

Answer:age of bone is 1756 yearsStep-by-step explanation:The amount of​ carbon-14 present in animal bones after t years [tex]P(t)= P_0e^{-0.00012t}[/tex]P(t) is the carbon present19% has lost. so carbon present is 100-19 = 81% presentout of 100 81 is presentsso P0 is 100P(t) is 81[tex]81= 100e^{-0.00012t}[/tex]divide by 100 on both sides[tex]0.81= e^{-0.00012t}[/tex]take ln on both sides[tex]ln(0.081)=-0.00012tln(e)[/tex][tex]ln(0.081)=-0.00012t[/tex]divide both sides by  -0.00012t=1756.0085age of bone is 1756 years