MHB Word problem-population with logs

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The population of UDAB in 1980 was 50,000, growing continuously at 5.5% per year, modeled by the formula P_U(t) = 5(1.055)^t. FILO's population started at 60,000, increasing by 3,000 annually, expressed as P_F(t) = 3t + 60. To find when UDAB's population reaches approximately 75,000, the equation 5(1.055)^t = 7.5 is used. Solving this leads to the logarithmic form t = log_1.055(3/2), which can be approximated using the change of base formula. The calculated time for UDAB to reach the target population is approximately 7.573 years.
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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!
 
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I think I would express the population of UDAB as (in tens of thousands):

$$P_{U}(t)=5(1.055)^t$$

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

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

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

$$P_{U}(t)=7.5$$

$$5(1.055)^t=7.5$$

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

$$5(1.055)^t=7.5$$

Divide through by 5:

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

Convert from exponential to logarithmic form:

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

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

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