Question

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?

Answer 1

`P=80e^0.015t` can be re-written as
`t=P/(80e^0.015)`

We are given that `P=120` (thousand).
So
`t = 120/(80e^0.015)`

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

Answer 2

Approximately 27 years.
I misread this question.

If the the model is: `P=80e^(0.015t)` with P measured in thousands, then:

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

We've been asked to solve:

`120=80e^(0.015t)`

`80e^(0.015t) = 120`

`e^(0.015t) = 120/80 = 3/2 = 1.5`

`e^(0.015t) = 1.5`

`0.015t = ln(1.5)`

`t= ln(1.5)/0.015`

Using a table of values or electronics gives an approximation of `t ~~ 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: `"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 `3/2` or `(5+sqrt7)/8)`