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?

#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

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)#