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Within an economic context, the concept of growth is extremely important. Extensive Growth is that which arises from adding new factors of production: be they workers, machines, or factories. This kind of growth is driven by capital accumulation or labour accumulation and exists in contrast to intensive growth, which occurs when existing factors are used more efficiently. An example of this is taking a bunch of separate machines and making them into an assembly line. No new machines are added, but the existing machines yield a higher output than before.

Growth can be charted using the Cobb-Douglass Production Function:

Y (Income) = T (Technology) * L (Labour) ^ (a) * K (Capital) ^ b where a+b=1

ln(Y) = ln(T) + a ln(L) + b ln(K)

Having a and b sum to one creates constant returns to scale within the model. In all developed economies, a is around .75 and b is around .25. Since income is allocated to factors according to their marginal contributions, it makes sense that 75% of national income (GDP) is directed to labour.

Within this model, changes in L and K represent extensive growth while changes in T represent intensive growth. In Canada, about 17% of GDP growth is attributable to changes in L, 13% to changes in K, and the remainder to changes in T.
Noded notes from Math Precalc class:

Equation to CalculateGrowth of a System Over Time

```YO=YIeKT
```
• YO = Amount of the object you will end with. (O is for Output)
• YI = Amount of the object you know you start with. (I is for Input)
• e = (see e for an explanation of the meaning of the term) On Texas Instruments graphing calculators, use the e^X button.
• K = Constant number in the equation.
• T = Time. When you are working in 1 year, T=1. When doing a problem monthly, T=12. Quarterly, T=4. Time does not have a specific unit. Therefore, the output number will be relative to the input value of T.

Theory Behind Solving Problems:
There are two scenarios nested inside each other in every question, a smaller case and a larger case. You are solving for the unknown in the larger case by way of the solvable smaller case. Use the smaller case to determine the constant for the purposes of that equation, and then use in the larger case to find the unknown there. If you are given K (the constant), then start by trying to solve the larger case.

Cull out all of your "given" data from the problem, and decide if it is for the smaller case or for the larger case. The smaller case needs a starting amount, an ending amount, and a time to get to that ending amount (i.e.- after X days, there will be Y rabbits in a cage). The larger case has a starting amount and either an ending time or an ending amount. One is a variable, the other has to be given.

Solve the smaller equation to find the constant for the problem (K).

Solve the larger equation using the K and the other ending amount or ending time.

Example Problem:

Ten "clan" of bacteria will multiply five times every twelve days. Given the duration of the month of April (30 days) to grow, how many bacteria "clans" will there be, to the nearest 10th of a clan?

What to do:

1. Find K (the constant) in terms of logs
To find the constant, the problem has to include three things: a starting amount of an object, a time, and how much of the object is left after that time.

Think about this before you start: In this problem, there are two times- twelve days, and thirty days. You need to use the twelve days, because you are told how many you have left after that point (five times the original).

From there, start putting pieces into the formula to solve for K. Remember your original equation:
YI=Y0eKT

Now sub in the beginning amount, 10 clans, for YI. This is culled straight from the question.
YO=10*eKT

Given that you start with 10 clans, you end up with five times the starting number. 10*5= 50. Sub 50 in as the ending amount in YO.
50=10*eKT

Then replace T with the time. Since you have two time amounts (12 days and 30 days), use the best unit between the two of them. Remember, since you are trying to solve the smaller case, see which correlates to an ending amount ("at 12 days, there is X left" or "at 30 days, there is X left"). In this problem, they are both days, so just make T in days. Replace T in the equation with 12.
50=10eK*12

Now you are down to one variable, K. This is what you need to solve for to find the constant in the equation. Use basic algebra and logarithm laws to solve the equation for K from here. First, divide out the 10 from both sides, and end up with:
5=eK*12

Then, take the logarithm of both sides (separately). When you take a log in the base of e, its called a natural log, and is written as ln e (should be written as script, lowercase LN). Since you need to take the natural log of the e side, you have to take the natural log of the 5 as well.
ln 5 = ln eK*12

Basic log law- when you have the log of a number to a power, it is the same as the log of the number times the power. Algebraically: log (XP) = P • log X . From that law, we can change the e side like so:
ln 5 = K • 12 • (ln e)

The natural log of e always equals 1. (another basic log law)
ln 5 = K • 12 • 1

Move the twelve over to the other side by dividing to get K alone.
```ln 5  = K
12```
Now that you have K alone, you know the constant for the larger equation that you have to solve.

2. Write an equation that includes the constant as a number instead of K
This portion requires a calculator that can manipulate e.
Go back to the equation at the top of the node that is all variables. We clear the slate, and start solving the larger equation from a "blank" equation. Then, plug in what we know. We know the starting amount is 10 clan:
YO=10•eKT

We also know that K = (ln 5)/12 from part A.
YO=10•e(ln 5)/12 • T

3. The variables we have left are Y and T. To get T, look back at the problem to find the time that you didn't use (30 days). Replace T with 30.
YO=10•e(30/12) • ln 5

Now there is only Y left to solve for. (For those that are wondering "what about e?!"- e is NOT a variable in this case. Think of it like π = 3.1415..., e=2.7... However, use a calculator for e, not 2.7. The decimal equivalent is not accurate enough for what we need.) Put the equation into your calculator as:
10e^((30/12)*ln(5))

And out comes 559.0169944. Round that off to the tenths (as the problem requires), and You have 559.0 clans after 30 days.
All math done on a TI-83+ calculator.

Growth (?), n. [Icel. gror, gri. See Grow.]

1.

The process of growing; the gradual increase of an animal or a vegetable body; the development from a seed, germ, or root, to full size or maturity; increase in size, number, frequency, strength, etc.; augmentation; advancement; production; prevalence or influence; as, the growth of trade; the growth of power; the growth of intemperance. Idle weeds are fast in growth.

Shak.

2.

That which has grown or is growing; anything produced; product; consequence; effect; result.

Nature multiplies her fertile growth. Milton.

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