In the series: Note 19.1.

Version: 0.1

By: Albert van der Sel

Doc. Number: Note 19.1: supplement to note 19..

For who: for beginners.

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Note 19 deals on exponential and logarithmic functions.

This note is a supplement to Note 19, and is a very quick intro

on exponential relationships.

-For more detailed information, it's best to refer to note 19.

-If you only need a very quick intro on the exponential relationship,

then this short note might be useful.

So, this is a stripped version of note 19, with a special focus on "the exponential relationship".

the graphs of y=3

The first one, is a steeply rising function, while the second one is a function which

rather quickly decends.

-If you consider y=f(x)=3

then you would indeed agree on the "shape" of that function.

Indeed: f(-1) = 3

-If you consider y=⅓

then you would indeed agree on the "shape" of that function.

Indeed: f(-1) = ⅓

Ofcourse, you may also see a function like y= c * a

which might be larger or smaller than "1". Ofcourse, when c=1, we simply have y= a

In the relation y=a

-If a < 1, the we have graphs like the y=⅓

as shown in the figre above. If "x"inceases, then the function decreases.

-If a > 1, we have graphs similar in shape as y=3

If "x"inceases, then the function increases as well, and the more steeply, the higher "a" is.

is used, but that one is completely equivalent to what we have seen above.

Since many researchers, study "something" which varies with "time", the "x" is often replaced by "t" (time).

Also, many studies relate a "number" or "amount" (like the amount or growth of bacteria) with time,

the usual x/y relationship is written as N/t relationship.

So, instead of:

-"b" is initial (starting) amount at t=0.

-"t" is time, similar to "x" in the convential notation.

-"N" is the new amount after "t" has passed, in which "b" has growthed (or declined) to "N".

-"g" is the base number, or also called the "growthfactor".

Instead of the usual XY coordinate system, with x and y values, we now have a relation

with N and t values, like if the x-axis is replaced by a t-axis. But fundamentally, nothing hs changed

at all.

where N (or N(t)) is the amount at a certain time (which varies with time), "b" is some initial number

at the starttime t=0, and "t" is time (similar as the role of "x" in the analytic description).

equation 2, works with any suitable unit of time. You might think that t in seconds might be

the preferred choice, but it simply depends on the subject under study.

Sometimes, it's more obvious to use days (like growth of the number of bacteria), and sometimes

it's more suitable to use years (like the acceptance of some social media among users).

Suppose we start with 10

Every day we measure the number of bacteria. Suppose that "g", the growthfactor, is 1.5.

Then we have:

at day 0: the starting "time": 10

at day 1: 10

at day 2: 10

at day 3: 10

etc..

Although N does not increase extremely fast, it's still an exponential relation, since

we have a constant

It's quite similar to the behavior of y=3

On day three we would have y=3

On day four we would have y=3

For every day "difference", we have constant factor of "3".

Let's consider an example of this way of doing business:

Suppose You have found the following data:

N |
Year |

30 |
2000 |

240 |
2001 |

1920 |
2002 |

15360 |
2003 |

122880 |
2004 |

So, what to make of this data? If the Number at each year, divided by the former number of the

year before that, produces a constant factor, then we are likely to have found

and exponential relation.

Let's try:

122880/15360=8, 15360/1920=8, 1920/240=8 etc..

It seems that we have found, per year, we have a constant factor of "8".

You may say that we have found a "growthfactor" of "8". Each year, the former N

has been multiplied with "8".

Looks like 8

Please notice, that this factor works

so it's not a linear relation.

If you would look at two years, like this example, 122880/1920= 64 = 8

You can also try something like that over 3 years, which gives us 8

The relation thus, smells like something which has a factor like 8

Also note that we started in the year 2000, up to 2004, meaning that we have 4 timesteps,

namely 2000->2001, 2001->2002, 2002->2003, 2003->2004.

Let's assume that our relation is:

we have started the measurements.

Would it work?. Let's find out. Using the formula, then over 4 years, the initial value "30"

must have grown to:

N = 30 * 8

Indeed, this matches the data of the table above.

This stuff decays into other producs, and suppose we have some sort of technique

to measure the remaining amount of that radioactive substance, over "t".

Suppose further that we have found the following measurements:

Amount |
At time: |

1000g |
t=0h |

700g |
t=1h |

490g |
t=2h |

343g |
t=3h |

240g |
t=4h |

Let's see how the amount have changed per hour:

240/343=0.7, 343/490=0.7 etc.. Here, I used the data over 3 hours.

It seems that our relation is:

N = 1000 * 0.7

That's in accordance with our measurements (see table), so our formula N = 1000 * 0.7

Suppose we have found an relation for some amount of "stuff", let's call that amount "N(t)",

which obeys the following relation:

N(t) = b * 0.95

where b is the initial value at some starting time, then the amount N(t) will decrease as time progresses.

We know exactly how much of the original stuff remains, since we have the expression to calculate it.

Practical applications are for example the halflife time for some radioactive substance.

Suppose that for some radioactive substance, the halflife τ is known. Then after τ time,

half of that radiactive material has decayed.

In the example N(t) = b * 0.95

½b = b * 0.95

(the ½ b is due to the fact that only half of the original amount remains).

So, in this case we must solve for "t" using 0.95

In general:

But if you would happen to know logarithmic functions, you would know that, for the example we used above:

0.95

So, this might help in finding concrete answers. If you are not so comfortable with logarithms, please see note 19.

This time, we might have a relation like for example N(t) = b * 1.3

growthfactor "g" > 1.

The Doubling time, is exactly that timespan, where the initial value "b" grows to 2b, so:

2b = b * 1.3

So in this example, you must solve "t" from: 1.3

In general:

But on most Highschools, it's common to use a graphical calculator if "g" is known,

so that you need to solve an equation like 1.3

If something growths or declines with a certain

express that as a growthfactor as well.

-Increase:

If some initial value increases with, say, with 17% per unit of time, then the associated growthfactor is:

So, we started out with 100% of the amount, and that grows with 17% per unit of time.

Thus, after the first time unit, we have 117% of the original amount.

This is equal to: b

This corresponds to 117% b.

Thus the general formula above, seems correct.

-Decrease:

If some initial value decreases with, say, with -17% per unit of time, then we can apply

the same formula, but this time we take into account the negative sign of that percentage.

So in the example of 17% decrease, we would have g = (1 - 17/100) = 0.83.

In the last example, your exponential formula would be: N=N

Note that sometimes I use "b" to denote te initial value (at t=0), and sometimes N=N

In case the initial value at t=0 would be, for example "1000", we would have N=1000 * 0.83

Thus, as usual, you can calculate the current number if you know the number of timeunits "t".

Without any proof, we state that: