# In how many years will the population of the city be 120,000 if the population P in thousands of a city can be modeled by the equation P=80e^0.015t , where t is the time in years?

Mar 29, 2015

$P = 80 {e}^{0.015} t$ can be re-written as
$t = \frac{P}{80 {e}^{0.015}}$

We are given that $P = 120$ (thousand).
So
$t = \frac{120}{80 {e}^{0.015}}$

with a little help from Excel:
$t = 2.216502$ years

Mar 29, 2015

Approximately 27 years.

If the the model is: $P = 80 {e}^{0.015 t}$ with P measured in thousands, then:

When the population is $120 , 000$, the variable $P$ will have a value of $120$

$120 = 80 {e}^{0.015 t}$

$80 {e}^{0.015 t} = 120$

${e}^{0.015 t} = \frac{120}{80} = \frac{3}{2} = 1.5$

${e}^{0.015 t} = 1.5$

$0.015 t = \ln \left(1.5\right)$

$t = \ln \frac{1.5}{0.015}$

Using a table of values or electronics gives an approximation of $t \approx 27.03$.

Notes:

1. Here I used $\ln$ to mean the natural logarithm function. If your teacher or textbook uses $\log$ to mean the natural logarithm, then you should too.
2. When solving an exponential function, it is most convenient to start by making the form of the equation: $\text{base"^"variable expression" = "constant expression}$. In this equation, that is the reason we divided by $80$ before taking the natural logs of both sides of the equation.
(A constant expression is a number, like $1.5$ or $\frac{3}{2}$ or (5+sqrt7)/8)