We are given that a population is modeled by the following function:
[tex]A=A_0e^{0.015t}[/tex]From this function, we can solve for the time "t". First, we divide both sides by A0:
[tex]\frac{A}{A_0}=e^{0.015t}[/tex]Now, we use the natural logarithm to both sides:
[tex]ln(\frac{A}{A_0})=ln(e^{0.015t})[/tex]Now, we use the following property of logarithms:
[tex]ln(x^y)=yln(x)[/tex]Applying the property we get:
[tex]ln(\frac{A}{A_0})=0.015tln(e)[/tex]We have the following:
[tex]ln(e)=1[/tex]Substituting we get:
[tex]ln(\frac{A}{A_0})=0.015t[/tex]Now, we divide both sides by 0.015:
[tex]\frac{1}{0.015}ln(\frac{A}{A_0})=t[/tex]Now, the time it takes for the population to go from "A = 5100" to "A = 6120" we need to subtract the final time from the initial time, like this:
[tex]\frac{1}{0.015}ln(\frac{6120}{A_0})-\frac{1}{0.015}ln(\frac{5100}{A_0})=t_f-t_0[/tex]Now, we take "1/0.015" as a common factor:
[tex]\frac{1}{0.015}(ln(\frac{6120}{A_0})-ln(\frac{5100}{A_0}))=t_f-t_0[/tex]Now, we use the following property of logarithms:
[tex]ln(\frac{x}{y})=ln(x)-ln(y)[/tex]Applying the property we get:
[tex]\frac{1}{0.015}(ln(6120)-ln(A_0)-ln(5100)+ln(A_0))=t_f-t_0[/tex]Now, we cancel out the "ln(A0)":
[tex]\frac{1}{0.015}(ln(6120)-ln(5100))=t_f-t_0[/tex]Solving the operations:
[tex]12.15=t_f-t_0[/tex]Therefore, the amount of time is 12.15 years.