Word problem-population with logs

[jibbs]

Hi, I've been staring at this problem for days and still cannot figure out how to begin:

In 1980 the population of UDAB was 50,000 people and since was increasing continuously by 5.5% per year for the next 30 years. On the other hand the population of FILO was 60,000 and increasing by 3,000 people per year over the same time period. For each country, write a formula expressing the population as a function of time, where t is the number of years since 1980. In how many years will the population of UDAB reach approximately 75,000 (use logs to solve).

I have in formula for UDAB as a=pert, A=50,000e^.055(30). I cannot figure out a formula for FILO. And while I can figure out how to solve the second part without logs, I am lost on how to do with using logs.

Any help would be greatly appreciated!

 

[MarkFL]

I think I would express the population of UDAB as (in tens of thousands):

\(\displaystyle P_{U}(t)=5(1.055)^t\)

And since the population of FILO increases by the same amount each year, I would write (the population in thousands):

\(\displaystyle P_{F}(t)=3t+60=3(t+20)\)

Then to answer the second part of the problem, we could write the equation:

\(\displaystyle P_{U}(t)=7.5\)

\(\displaystyle 5(1.055)^t=7.5\)

Now, solve for $t$...what do you find?

 

[MarkFL]

We have:

\(\displaystyle 5(1.055)^t=7.5\)

Divide through by 5:

\(\displaystyle (1.055)^t=\frac{3}{2}\)

Convert from exponential to logarithmic form:

\(\displaystyle t=\log_{1.055}\left(\frac{3}{2}\right)\)

Use change of base formula to get an expression we can use to obtain a decimal approximation using a calculator:

\(\displaystyle t=\frac{\ln\left(\dfrac{3}{2}\right)}{\ln(1.055)}\approx7.573\)