Function relating to population growth

[astral_stardust]

I'm stuck and could use a hint, or someone who understands it better than I do explaining what the function means. I'm not too good with setting up word problems.

Show that for any positive constants a and k, the function g(t)=a2^kt doubles after 1/k years. t represents time in years and the function represents population of a country as a function of time.

 

[stapel]

I'm stuck and could use a hint, or someone who understands it better than I do explaining what the function means. I'm not too good with setting up word problems.

Show that for any positive constants a and k, the function g(t)=a2^kt doubles after 1/k years. t represents time in years and the function represents population of a country as a function of time.

What are your thoughts? What have you tried? How far have you gotten? Where are you stuck?

For instance, they claim that, regardless of the values of "a" and of "k", the given equation will give some value for whatever value you plugged in for the time "t", and will give a doubled value for a t-value that is 1/k more than the initial t-value. So you plugged "t = 0" in, and got... what? Then you plugged "t = 1/k", and got... what? You plugged in "t = 1" and got... what? You then plugged in "t = 1 + 1/k" and got... what? What did you conclude?

Please be complete. Thank you!

 

[Ishuda]

I'm stuck and could use a hint, or someone who understands it better than I do explaining what the function means. I'm not too good with setting up word problems.

Show that for any positive constants a and k, the function g(t)=a2^kt doubles after 1/k years. t represents time in years and the function represents population of a country as a function of time.

A slightly different way that what stapel has:
\(\displaystyle \frac{\text{Value at t + }\, \frac{1}{k} \text{ years}}{\text{Value at t years}}\)\(\displaystyle \, =\, \frac{a\, 2^{k\, (t\, +\, 1/k)}}{a\, 2^{kt}}\) = ?

 

[Dale10101]

Understanding

I'm stuck and could use a hint, or someone who understands it better than I do explaining what the function means. I'm not too good with setting up word problems.

Show that for any positive constants a and k, the function g(t)=a2^kt doubles after 1/k years. t represents time in years and the function represents population of a country as a function of time.

You are asking for a better understanding of what the function means, this might prompt some ideas of your own.

let a = 1, and k = 1

then we have are left with "understanding" what the function 2^t means,
Now suppose t takes on the values of 1,2 and 3 successively:

2^1 = (2)
2^2 = (2)(2)
2^3 = (2)(2)(2)

you can see that each time you increment t by an additional year, you add another doubling factor, (2) to the string of factors that 2^t represents. In a sense you could say that 2^t counts by doubles.

Now suppose k = 3 then for successive values of t = 1,2 and 3:

2^{[(3)](1)} = 2^3 = {[(2)(2)(2)]}, i.e. one factor of {[(2)(2)(2)]} for t = 1

2^{[(3)](2)} = {[(2)(2)(2)]}{[(2)(2)(2)]} i.e two factors of {[(2)(2)(2)]} for t = 2

2^{[(3)](3)} = {[(2)(2)(2)]}{[(2)(2)(2)]}{[(2)(2)(2)]} i.e three factors of {[(3)(3)(3)]} for t = 3

In this case when you increment t by one year you add an additional factor of (2^3) = (8) to the factor string. You might say that you are counting by multiples of 8.

In short, to a first order, you can think of a factor raised to a power as the number you get by evaluating a product string of factors whose length is equal to the power that the factor is being raised to.

For example,

m^t = (m)(m)(m) …. (m), a product sting with “t” number of factors of m.

Also,

m^(kt) = (m)(m)(m) … (m), a product sting whose length is “kt” number of factors of m,

BUT … by a law of exponents m^(kt) = (m^k)^t so you can also think of m^(kt) as the factor sting,

(m^k) (m^k) m^k) … (m^k), with “t” number of factors of (m^k).

Now the problem’s author claims that if you evaluate the function n = (a) 2^(kt) for any k and t, for successive values of 1/k years, n will double. This claim means that in 1/k years, n will double, and in another 1/k years the last result will double again, and in another 1/k years THAT last result will double again, and so on. So putting this to the test:

n₁ = (a)(2)^[(k)(t)] = (a)(2)^[(k)(1/k)] = (a)(2)^1 = 2a, so at the end of the first 1/k interval “a” has doubled

For the second interval of 1/k years we face the same problem except that at the beginning of the second interval our starting value is 2a, since “a” doubled in the first 1/k years.

n_{2 =}(2a)(2)^[(k)(t)] = (2a)(2)^[(k)(1/k)] = (2a)(2)^1 = 4a, so again, in an interval of 1/k years, the value of n doubles.

Repeating this exercise a third time with the same result constitutes the demonstration of an apparent pattern and THAT might be enough for “showing” the validity of the claim presented. To go the final step to “proving” the claim one might use the “Principle of Mathematical Induction”.

Of course Ishuda has demonstrated a much simpler method of proving that the ratio of a next result to a previous result is a factor of 2, i.e a doubling.

a2^k(t+1/k) / a2^kt​ = 2

Hope this helps with the “understanding” part of your question.