We are given function: [tex]g(x)= 112\ ln (0.121x) + 2011[/tex].
Given function models a particular year of population of New York City.
x represents population of New York City ( In millions).
We need to estimate the population of New York City in 2020.
Because g(x) function represents a particular year of population of New York City and we are given year 2020, so we need to replace g(x) by 2020 and solve for x.
Replacing g(x) by 2020, we get
[tex]2020= 112\ ln (0.121x) + 2011[/tex] : This is the required equation to estimate the population of New York City in 2020
Let us solve the above equation for x now.
[tex]2020=112\ln \left(0.121x\right)+2011[/tex]
[tex]\mathrm{Switch\:sides}[/tex]
[tex]112\ln \left(0.121x\right)+2011=2020[/tex]
[tex]\mathrm{Subtract\:}2011\mathrm{\:from\:both\:sides}[/tex]
[tex]112\ln \left(0.121x\right)+2011-2011=2020-2011[/tex]
[tex]Simplify[/tex]
[tex]112\ln \left(0.121x\right)=9[/tex]
[tex]\mathrm{Divide\:both\:sides\:by\:}112[/tex]
[tex]\frac{112\ln \left(0.121x\right)}{112}=\frac{9}{112}[/tex]
[tex]\mathrm{Simplify}[/tex]
[tex]\ln \left(0.121x\right)=\frac{9}{112}[/tex]
[tex]\mathrm{Apply\:log\:rule}:\quad \:a=\log _b\left(b^a\right)[/tex]
[tex]\frac{9}{112}=\ln \left(e^{\frac{9}{112}}\right)[/tex]
[tex]\ln \left(0.121x\right)=\ln \left(e^{\frac{9}{112}}\right)[/tex]
[tex]0.121x=e^{\frac{9}{112}}[/tex]
[tex]\mathrm{Divide\:both\:sides\:by\:}0.121[/tex]
[tex]\frac{0.121x}{0.121}=\frac{e^{\frac{9}{112}} }{0.121}[/tex]
[tex]x=8.95598[/tex] ≈ 9 million people.
So, we can say..
Population of New York City in 2020 would be 8.95598 millions or 9 million ( approximately).
