Rates & Work Strategy: Population Prblms

[mozart611]

I tried to solve #4 pg 45 in Word Translations using the equation for exponential growth, but I was unable to. I thought the equation was a faster approach. Here's the question
The population of grasshoppers dbles in a field every year. Apprxly how many yrs will it take to grow from 2,000 to 1,000,000 grasshoppers or more

Eqn for Exp growth: P = S.2^(t/I)

1,000,000 = 2,000. 2^(t/2)

Thnks

[StaceyKoprince]

I do actually like just writing these out (the way it's done in the book). You don't even have to write all the zeroes: just know that you're trying to go from 2k to 1,000k and write that out without those last three zeroes. If it can be done this way, this is usually the easiest (read: least likely to make a mistake) way to do the problem.

This is especially true if the question asks you to approximate and / or asks you to solve for not one specific number but "some number or more / less." (Both of these conditions are present in the given problem.)

The generic equation for a doubling population is:

Pop = S*2^(t/I)
where:
Pop = final population
S = starting population
t = time amount of time it takes (what we're looking for)
I = the interval (the amount of time it takes to double)

In this case:
Pop = 1,000,000
S = 2,000
t = unknown
I = 1 year

You have I = 2 in your equation.

So, here's the filled-in equation:
1,000,000 = 2,000*2^(t/2)

And now you can see the problem with using this equation to solve. How do you get that t out of the exponent? The value for t is almost certainly not going to be an integer (otherwise, they wouldn't have said "approximately" and "1,000,000 or more"). The only real way to solve exactly for t here would be to use logs... and you don't have a calculator.

So skip the formula - just use the chart. :)